
Class _^A ^'1 
Book , £-3T v 



GopigM . 



COPYRIGHT DEPOSIT. 



AN INTRODUCTION 

TO 

PROJECTIVE GEOMETRY 



AND ITS 



APPLICATIONS 



AN ANALYTIC AND SYNTHETIC 
TREATMENT 



BY 

ARNOLD EMCH, Ph.D., 

Professor of Graphics and Mathematics in the University of Colorado 



FIRST EDITION 
FIRST THOUSAND 



NEW YORK 

JOHN WILEY & SONS 

London: CHAPMAN & HALL, Limited 

*9°5 



- 

.B ' 



LIBRARY of CONGRESS 
Two Copies Received 

JAN 5 1905 

Copyrurnt entry 

CUSS Sf XXc Not 

COPY B. 



Copyright, 1905, 

BY 

ARNOLD EMCH. 



ROBERT DRUMMOND, PRINTER, NEW YORK. 



PREFACE. 



Treatises on Projective Geometry are usually written with 
the object of presenting this science in a purely systematic form; 
hardly any attention is paid to the applications. As a rule the 
methods of " arithmetized " mathematics have to be transformed, 
made more concrete, before they lend themselves to the solution 
of practical problems; and this, in the judgment of many, dis- 
figures the purely scientific method. 

In this respect, projective geometry, geometry of position, is 
no exception. The puristic tendencies of von Staudt, Reye, and 
others culminate in the modern Italian school of geometer-logicians, 
headed by Veronese 1 and Enriques. The latter' s projective 
geometry 2 contains an admirable logical presentation of the 
subject. With Enriques projective geometry is a ""visual" sci- 
ence, and everything is foreign to it which cannot be based upon 
the axioms of vision. 

It seems doubtful whether the axioms of vision alone can 
establish a sound projective geometry. Enriques himself, in 
his book, lets the fundamental elements of the first order be 
generated by motion! In this visual geometry metrical proper- 
ties, which are indispensable in the applications, appear as special 
cases and are of secondary importance. Conies result from the 
theory of polarity. 

On the other hand, Fiedler, Wiener and others show that 
the methods of Poncelet, Steiner, Chasles, and Cremona naturally 



1 Grundziige der Geometrie, Teubner, Leipzig. 

2 German translation, Teubner, Leipzig. 



iii 



IV PREFACE. 

follow from the study of descriptive geometry. With them, 
projective and descriptive geometry are organically related and 
each branch benefits by its connection with the other. Little 
attention is paid to the so-called foundations. 

As the present book has been written with a utilitarian pur- 
pose, considerable space is given to the applications; and in 
their treatment use has sometimes been made of original analytic 
and geometric methods of attack and solution. It has thus been 
found possible to include some new subject-matter and especially 
certain parts of modern analytical geometry. 

In addition to the traditional contents of the standard ele- 
mentary treatises, two chapters on pencils and ranges of conies, 
including cubics, and on the applications in mechanics have 
been added. The Steinerian transformation contained in Chap- 
ter IV, in connection with the study of plane cubics, is a brilliant 
example of the original power of projective geometry; and as it 
is elementary, it seems natural to introduce it after the theory of 
conies. As a novel feature the realization of collineations by 
linkages, described in Chapter V, may be mentioned. 

Much time may profitably be devoted to the original prob- 
lems and to the constructions involved in them. No first study 
of projective geometry can be successful without the constant 
use of ruler and compass. 

My thanks are due to my colleague, Professor Ira M. De Long, 
for many valuable suggestions as to matters of form. 

Corrections and suggestions as to either the form or the matter 
of the text are earnestly solicited. 

Arnold Emch. 
Boulder, Colorado, 
July, 1904. 



CONTENTS. 



CHAPTER I. 



General Considerations. Anharmonic Ratio. Projective Ranges 
and Pencils. Polar Involution of the Circle. 

PAGE 

§ i. Geometric quantities and their signs i 

§ 2. Anharmonic ratio. Projective transformation of the points of a 

straight line 5 

§ 3. Involution n 

§ 4. Projective pencils of rays 15 

§ 5. Involutoric pencils 18 

§ 6. Product of projective pencils and ranges 22 

§ 7. Exercises and problems 25 

§ 8. The complete quadrilateral r6 

§ 9. Perspective pencils and ranges 29 

§ 10. General construction of projective pencils and ranges 30 

§ 1 1. Exercises and problems 34 

§ 12. Projective properties of the circle 35 

§ 13. Polar involution of the circle 38 

§ 14. Continuation of § 13 41 

CHAPTER II. 

COLLINEATION. 

§ 15. Central projection 45 

§ 16. Analytical representation of central projection ; . 49 

§17. Special cases of central projection 51 

§ 18. Exercises and problems 57 

§ 19. Collineation 59 

§ 20. Geometrical determination and discussion of collineation 63 

v 



vi CONTENTS. 



PAGE 



§21. Continuous groups of projective transformations 66 

§ 22. The principle of duality 68 

§ 23. Exercises and problems 70 

§ 24. Orthographic projection 72 

^ 25. Affinity between horizontal and vertical projections of a plane 

figure 77 

^ 26. Homologous triangles 80 

I 27. A few applications of perspective 86 

J 28. Exercises and problems , 90 

CHAPTER III. 

Theory or Conics. 

) 29. Introduction 92 

j 30. Identity of curves of the second order and class with conics 93 

\ 31. Linear transformation of a curve of the second order 94 

\ 32. Polar involution of conics. Center. Diameters. Axes. Asymp 

totes 102 

\ 33. Existence of ellipse, hyperbola, parabola, and their foci 104 

\ 34. Construction of foci independent of central projection 107 

) 35. Focal properties of conics 109 

i 36. Analytical expression for tangents and polars. Equation in line- 
coordinates 118 

\ 37. Theory of reciprocal polars 123 

i 38. General reciprocal transformation. Polar systems 126 

i 39. Theorems of Pascal and Brianchon 133 

i 40. Applications of Pascal's and Brianchon's theorems 137 

i 41. Conics in mechanical drawing and perspective 140 

i 42. Special constructions of conics by central projection and parallel 

projection 146 

t 43. Problems of the second order 161 

i 44. An optical problem 167 

CHAPTER IV. 

Pencils and Ranges of Conics. The Steinerian Transformation. 

Cubics. 

t 45. Pencils and ranges of conics 172 

; 46. Products of pencils and ranges of conics 180 

( 47. The Steinerian transformation 185 



CONTENTS. vn 

PAGE 

t 48. Curves of the third order 189 

t 49. Curves of the third order generated by involutoric pencils 197 

i 50. Various methods of generating a circular cubic 204 

51. The five types of cubics in the Steinerian transformation 209 

CHAPTER V. 

Applications in Mechanics. 

52. A problem in graphic statics 217 

53. Statical proofs of some projective theorems 220 

Geometry of stresses ln a plane 223 

54. General remarks 223 

55. Involution of conjugate sections and stresses 224 

56. Discussion of this involution 227 

57. The stress ellipse. Metric properties of the involution of stresses . . 229 

58. Examples . 236 

59. The rectangular pair of the involution of stresses in nature 239 

Realization of colllneations by linkages. 242 

60. Introductory remarks 242 

61. Analytical formulation of the problem 244 

62. Peaucellier's inversor 246 

63. Pantographs 248 

64. Rotator and its combinations 251 

65. Translators , 252 

66. Linear transformation 253 

67. Perspective 257 

Index 261 



PROJECTIVE GEOMETRY. 



CHAPTER I. 



GENERAL CONSIDERATIONS. ANHARMONIC RATIO. PROJEC- 
TIVE RANGES AND PENCILS. POLAR INVOLUTION OF THE 
CIRCLE. 

§ i. Geometric Quantities and their Signs. 

Geometric quantities can be represented by numbers by 
assuming an arbitrary geometric quantity of the same kind as a 
unit. 1 To show this for linear quantities, assume any line AZ, 
Fig. i, and a unit u. Measure off on AZ as many units u as 

A B Z 

* 1 1 1 1 1 1 

u ~ 

Fig. i. 

possible, so that the remainder BZ<u. Suppose that the num- 
ber of units measured on A Z is a, so that AZ=au-\-BZ. Now 
consider BZ as a unit and u as the quantity to be measured. 
Suppose that BZ is contained b times in u and that the remain- 
der r<BZ. Then u = b-BZ+r. In a similar manner, consider r 
as a unit and BZ as the quantity to be measured. Suppose that 
r is contained c times in BZ and that the remainder is s, so 

1 See Lagrange's Lectures on Elementary Mathematics (translated by Th. J. 
McCormack, Open Court Publ. Comp., Chicago), p. 3. 



2 PROJECTIVE GEOMETRY. 

that BZ=cr+s. Continuing this process till it closes, or else 
indefinitely, there results the series 

AZ = au+BZ, 
u=bBZ+r, 

BZ = cr+s, 

r=ds+t, 



which by elimination leads to the continued fraction (u as the 
original unit being i) 

AZ=a+i 



b+i 



c+i 



d+i 

6+ . .'. 

This evidently represents a number and the proposition is 
proved. 1 

Geometric quantities being represented by numbers, it must be 
possible to define negative and positive geometric quantities in 
accordance with the laws of Arithmetic. This can be done in the 
most convenient manner by the method of displacements. As- 
sume any line and three points on it in the order A, B, C from 

1 This continued fraction is convergent, since by the process of its formation 
it continually approaches the limit, which is known in advance. For example, 

2+1 

2+ . . . 

Concerning further details consult Laurent: Traite d' Analyse, Vol. V, pp. 321- 

359- 



GENERAL CONSIDERATIONS. 3 

the left to the right. Increase and decrease of geometric quan- 
tity on this line are measured by the amount of displacement of a 
moving point, or by the length of the line between the original 
and final position of a moving point on this line. The formal 
laws of all displacements on this line are those of the group. Thus, 

(i) AB+BC=AC 

shows that two displacements succeeding each other are equiva- 
lent to a single displacement of the same kind and of the same 
system (group). 1 It follows further that 

(2) AB+BC+CA=o\ 2 

hence by substitution of (1) in (2) 

AC+CA=o, 
or CA = -AC; 

i.e., two displacements, or geometric quantities, which are de- 
scribed in opposite directions are of opposite sign. The same 
conclusions are reached when angular displacements are con- 
sidered. It is a universal -convention to designate all geometrical 
quantities which are obtained by displacements on a line from 
the left to the right as positive and those in the opposite direction 
as negative. In a similar manner, angles formed by angular dis- 
placements counter-clockwise are assumed as positive and those 
clockwise as negative. The conception of the group is general 
and also comprises the determination of such geometric quan- 
tities as areas and volumes. 

In case of a surface assume a pole O and any line / on this 
surface. A point P is moving on /, and in any position of its 
motion is connected to the point O by a geodesic of the surface. 
The generalized radius vector OP then sweeps over a certain area 

1 It is beyond the limits of this book to enter into a discussion of groups in 
this connection. 

2 M6bius: Barycentrische Calcul, § 1. 



PROJECTIVE GEOMETRY. 



which is subject to the laws of the group. Thus if P moves 
from A to B to C, counter-clockwise with respect to O, 

OAB+OBC = OAC, 
OAB+OBC+OCA=o; 



hence 



or 



OAC+OCA=o, 

OCA =-OAC. 



Two areas have therefore to be considered as of opposite sign 
if their boundaries are described in oppo- 
site senses. According to the distinc- 
tion which we have made, an area is 
positive or negative according as its 
boundary is described counter-clockwise 
or clockwise, respectively. From this it 
follows further that if the point P de- 
scribes a closed line on a surface, the 
radius vector OP sweeps over an area 
equal to the area enclosed by the curve. 
For a triangle ABC, Fig. 2, 

ABC = OAB+OBC+OCA, 




Fig. 2. 



where OCA is negative. On the other hand 

CBA = OCB + OB A + OAC 

= -OBC-OAB+OAC; 

hence ABC+CBA=o, 

CBA = - ABC. 

The same reasoning may be extended to the determination of 
volumes, which is left to the reader as an exercise. 

Ex. 1. If A, B, C, D are four collinear points, prove that 

BC-AD+CA BD + AB-CD = o. 



GENERAL CONSIDERATIONS. 5 

Ex. 2. For the same points prove 

DA 2 -BC+DB 2 -CA + DC 2 'AB=-BC-CA'AB. 

§ 2. Anharmonic Ratio. 1 Projective Transformation of the 
Points of a Straight Line. 

i. Critical Note. — von Staudt in his classical works 2 on the 
geometry of position created a system with the principal purpose 
of laying the foundations of geometry without the aid of metrical 
considerations. He introduced the word "Wurf" as an equiva- 
lent of anharmonic ratio and attached to it a meaning independent 
of any ratio. The anharmonic ratio is considered as a property 
of the "Wurf", so that, according to v. Staudt, metrical geometry 
is based upon projective geometry, or rather the geometry of 
position. Steiner, on the contrary, took the anharmonic ratio as 
a starting-point in his investigations. 3 In a recent paper 4 Poin- 
care has pointed out "that from a certain point of view the 
geometry of v. Staudt is predominantly a visual geometry, while 
that of Euclid is predominantly muscular." In other words, the 
two geometries are derived from experiences in optics and kine- 
matics, respectively. 

In works with practical purposes, where applications form an 
important part, it is probably of the greatest advantage to take 
one view or the other according to the simplicity of the treatment 
which it may afford. 

This method, although objectionable from the standpoint of 
pure geometry, reflects the development of geometric science 
itself. 



1 I shall use the expression anharmonic ratio, because it is used by the trans- 
lators of Reye's and Cremona's treatises on projective geometry and by a majority 
of English authors. Double ratio, corresponding to the German Doppelverhalt- 
nis, is presumably a better designation. 

2 Geometrie der Lage, 1847. Beitrdge, 1856-60. 

3 Systematische Entwickelung der Abhangigkeit geometrischer Gestalten, etc., 
1832. 

4 On the Foundations of Geometry, Monist, No. 1, Vol. IX. 



6 PROJECTIVE GEOMETRY. 

2. The anharmonic ratio of four points A, B, C, D on a line, 






A 


B 


C D 






1 


! ! 


j< 


1 




M d "*• 



Fig. 3. 



a straight line for the sake of simplicity, where (A, B) shall be 
designated as the first, (C, D) as the second pair, Fig. 3, is 



(1) 



AC 1 AD 



BC I BD 



k. 



As AC, BC, AD, BD are all positive quantities, k will be a posi- 
tive number. It is clear that this is not the only anharmonic 
ratio that may be formed between the four points. As there are 
24 permutations possible between four elements, there will also 
be 24 anharmonic ratios. Some of these, however, have the 
same value, and it may easily be verified that there are only 6 
different anharmonic ratios possible. Designating (1) by the 
symbol (A BCD), 1 these are 

(ABCD) = k, 



(2) 



(BACD)=j 
(BCAD) = 
(CBAD) = 



-k 
k ' 
k 



(CABD)=— k , 
(ACBD) = i-k. 



If the points A, B, C, D are located by their displacements a, b, 



Mobius, Barycentrische Calcul, § 183. 



PROJECTIVE TRANSFORMATION . 7 

c, d from a fixed point O, the first anharmonic ratio assumes the 
form 

c— a id— a 

(3) ^i/d^b =k - 

3. This expression leads to the solution of the important prob- 
lem to find all pairs of points, X, F, which with two fixed points 
A and B form the constant anharmonic ratio k. Associating with 
X and F the displacements x and y from O, the condition, ac- 
cording to (3), is 

x— a ,y—a 

(4) xrr7T=* 



or, solved for x, 

(s) 



x—b y—b ' 

(a— bk)y— ab(i — k) 
(i—k)y—(b—ak) 



From this it is seen that to every value of y corresponds one 
and only one value of x satisfying the condition of the problem, 
and vice versa. Taking any four points Y 1} Y 2 , F 3 , F 4 , and de- 
termining the corresponding points X v X 2f X 3J X 4 according to 
(5), there is found the relation 

(X 1 X 2 Z 3 X 4 ) = (F 1 F 2 F 3 F 4 ); 

i.e., any four points of the series (X) and their corresponding 
points of the series (F) satisfying the condition (5) have the same 
anharmonic ratio. Two series or ranges of points with this 
property are said to be projective. Formula (5) is the analytical 
expression for these projective ranges of points; it effects a pro- 
jective transformation 1 of the points of a straight line. 

For y = a, x = a, and for y = b, x=b\ i.e., the transformation 
leaves the points A and B invariant; they are called the double- 

1 The word projective was first used by Poncelet in his great work : Traite 
des proprietes projectives des figures, 1822. Mobitjs was the first who gave an 
analytical representation of projective transformations, in Der barycentrische 
Calcul, 1827. 



PROJECTIVE GEOMETRY. 



points of the transformation, or of the projective ranges of points. 
From (4) follows immediately that every pair oj corresponding 
points jornis a constant anharmonic ratio with the double- points. 
On the other hand every transformation of the form 



(6) 



X 



Ay+B 
Cy+D 



is projective. To prove this assume four points Y ly Y 2 , Y 3 , Y 4j 
and determine the corresponding points X lt X 2 , X 3 , X 4 . Let 
Vf, y 2 , y 3 , y 4 and x ly x 2 , x 3 , x 4 be the corresponding displacements, 
then to form (X t X 2 X 3 X 4 ) we have from (6) 

(AD-BC)(y 3 - yi ) 



(7) 









vVo vV« 



4 1 



Ji/A •X'n 



(Cy 3 +D)(C yi +D)> 

JAD-BQ(y t -y t ) 

(Cy 3 +D)(Cy 2 +D)' 

(AD-BC)(y 4 - yi ) 
(Cy t +D)(C yi +D)> 

(AD~BC)(y ( -y 2 ) 
(C yi +D)(Cy 2 +D)' 



and by division 









■ 1 / x A -x 1 _y 3 -y l /y 4 



oc±-x 2 y 3 -y 2 i y*-y 2 
or (X 1 X 2 X 3 X 4 ) = (Y 1 Y 2 Y 3 Y 4 ), 

which is a property of a projective transformation. To prove 
that (6) is of the form (5), we find the double-points of (6) by 
putting y = x; then (6) becomes 

(8) Cx 2 -(A-D)x-B = o; 

hence, designating the roots of this equation by a and b, 

A-D + V(A-D) 2 +4BC 



(9) 



2C 

A-D-x/(A-B) 2 + 4 BC 
2C 



PROJECTIVE TRANSFORMATION. 9 

The transformation (6) has therefore two double-points. Put- 
ting in the first and third equations of (7) x 3 = y 3 = a and 
x 4 = y i = b, it is found by division that 

a~x 1 /b-Xi Cb + D 

(10) /t =- — — r = & (constant). 

v J &—yJ b—y 1 Ca+D v ' 

Thus we find that any pair of corresponding points of the 
transformation (6) forms a constant anharmonic ratio with its 
double-points; such a transformation is projective. In deriving 
equation (5) it was assumed that A and B are real points. 
Assuming a projective transformation of the form (6), where 
A, B, C, D are real coefficients, it may happen that the double- 
points given by (9) are imaginary. In fact there are three pos- 
sibilities for the double points. According as 



(11) (X-P) 2 +4#c|o, 



a and b, or the double-points, are real, real and coincident, or 
imaginary, and the transformations are then called hyperbolic, 
parabolic, or elliptic. 

4. We shall next show that two projective ranges are deter- 
mined by three pairs of corresponding points X ± , Y x ; X 2 , Y 2 ; X s , Y 3 , 
whose positions are determined by the coordinates x 1} y ± ; x 2 , y 2 , . . . 
If these points are corresponding in two projective ranges, their 
coordinates must satisfy some relation of the form 

ay+b 
cy-\-a 

or cxy+dx—ay—b = o. 

c d a . 

To determine the ratios 7-, 7-, -r, which evidently determine the 

transformation, we have the conditions 



IO 



PROJECTIVE GEOMETRY. 



c da 

■^i+jai-p'i-i-o, 
c da 



c da 

These are three equations with the three required ratios as 
unknown quantities. These are therefore uniformly deter- 
mined by the x's and y's and are in determinant form: 



I 


X, 


Jl 


I 


x 2 


y 2 


I 


x 3 


y* 


*&% 


x x 


y± 


^2^2 


X 2 


y 2 


x 3 y 3 


x 3 


ys 



x& 


I 


yi 


x 2 y 2 


I 


y 2 


^3^3 


I 


y s 


x$ x 


x ± 


yi 


#2^2 


x 2 


y 2 


^3 


x 3 


y 3 



*Wi 


-*1 


I 


^2^2 


~x 2 


I 


*tfi 


%3 


I 



x x y x 

^3^3 



x, 



This proposition is also geometrically clear. In two pro- 
jective ranges any four points of one range have the same anhar- 
monic ratio as the four corresponding points of the other range. 
Hence, choosing any fourth point x 4 , then there is clearly only 
one point F 4 , so that 

{X 1 X 2 X z X i ) = (Y 1 Y 2 Y 3 Y i ); 

i.e., three pairs of points determine the projectivity. 

As an exercise assume the case of two coincident projective 
ranges for which the infinitely distant point is self-correspond- 
ing. Let x 1 = y 1 determine this infinitely distant point. From 

c .da 

the above expressions we find t~ = o, while j- and j- are finite. 

The projective transformation assumes the form 

x=ay-r-(3; 
i.e., what is called a linear transformation. 



INVOLUTION. II 

5. It is beyond the limit of this book to discuss all special 
cases of projective transformations of a straight line in detail. 
We shall indicate one of its properties which is of extreme impor- 
tance in modern geometry, and then discuss the special case of 
involution. Let a point x be transformed into a point x! by 
the projective transformation 

ax-\-B 

Transform x' into a point x" by another transformation of 
the same kind: 

(13) *"=- /T> - 

The result of these two operations is 

which shows that x" is obtained from x by a projective trans- 
formation of the form (12). Hence one, two, or more opera- 
tions of the form (12) in succession are equivalent to an oper- 
ation of the same kind. Giving a, /?, j-, d all possible real values, 

Oi B y 

(12) depends upon the three ratios-*-, -j , -*, so that (12) repre- 
sents a triply infinite number of projective transformations. 
For this reason it is said that all projective transformations of a 
straight line form a continuous three-termed group (dreigliedrig). 1 



§ 3. Involution. 2 

In case that the constant anharmonic ratio k in equation (4) 
of the foregoing paragraph is — 1, 

1 Sophus Lie: Vorlesungen liber continuierliche Gruppen. 

2 First systematically studied by Desargues (Brouillon projet, etc.). 



12 PROJECTIVE GEOMETRY. 

x—a y—a 
v J x—by—b 

(a+b)y—2ab 

(2) *"" 2y-(a+b)' 

In these equations x and y can be interchanged without affect- 
ing (i) or (2). The ratio (ABXY) = — 1 is called a harmonic ratio 
and (1) and (2) represent an involutoric transformation. To the 

a+b 
point at infinity, , y = 00 , corresponds the point x= ; i.e., 

the point bisecting the distance AB between the double-points. 
It is called the middle point of the involution. Designating this 
point by M, it is found that 

(3) ' MX.MyJ^ = ^; 

y ^ J 44' 

i.e., the product 0) the displacements 0} two corresponding points 0} 
an involution from the middle point is constant and equals the 
square of the displacement of either double-point from the middle 
point. 

Equation (2) may always be written in the form 

<4) x = °^^, 

if a and b are given by the values 

a la 2 B . a la 2 /? 

r x r 2 r r V r 

As these expressions define the double-points, they must also 
result directly from (4). For the double-points x=-y\ hence 
from (4) 

yx 2 — 2<xx+(3 = o. 



INVOLUTION. 1 3 

The roots of this equation are indeed identical with the pre- 

B a 2 
vious values of a and b. If - >— , a and b are conjugate complex 

numbers, i.e., the double-points of the involution are imaginary. 
In this case the middle point M of the involution is still real, 

a+b a 

since = - , and 

2 r > 



MX-MY = 



(a-b)' 



a 2 B 

— --<o; 

f r 



B a 2 
X and Y are on different sides of M. For - = — » the double- 

r r 

points coincide, and MX i MY=o\ every point corresponds to M. 
According to these results involution has been classified as hyper- 




Fig. 4. 

bolic in case of real double-points, elliptic in case of imaginary 
double-points, parabolic in case of coinciding double-points. 




Fig. 5. 

Geometrically, the different cases may be obtained as intersec- 
tions of a straight line with coaxial" systems of circles. Figs. 4, 



14 



PROJECTIVE GEOMETRY. 



5, and 6 represent hyperbolic, elliptic, and parabolic involutions 
respectively. In the first the points of a pair, XY, are always on 
the same side of M, and move in opposite directions ; in the second 
they are on different sides of M, and one of the points (X) is 




Fig. 6. 



within the distance AB and the other without. Corresponding 
points move in the same direction. We have seen that an in- 
volution on a straight line is determined by the transformation 



(5) 



ax+b 



ex— a 



or 



(6) 



exx' — a(x+x') — b = o, 



which shows that an involution is determined by two pairs, since 
there are only two essential constants in (6). 
Suppose that in a projective transformation 



, ax+b 
x f = — — -,, or 
cx+d 



cxx f + dx' — ax— b = o y 

the points x/, x± may be interchanged without affecting the pro- 
jectivity. The condition for this is 



(7) 
(8) 



cx x x x ' + dx/— ax x — b = o, 
cx x x{ + dx x — ax/ —b = o. 



PENCILS AND RANGES. 15 

Ey subtraction 

(9) d(x x - #/) + a(x t — x l ')=o, 

which can only be satisfied when d= — a, since x^xj. The 
condition d= — a, however, implies involution, hence the theorem: 
// a projective transformation contains a pair whose values may 
be interchanged without altering the transformation, it is an involu- 
tion. Thus if x x x{ be a pair, 

(10) (ABX.X^) =ABX 1 'X 1 ) = - 1. 

§ 4. Projective Pencils of Rays. 

Let a, b, c, d be four rays (straight lines) passing through 
a common point, and (rib), (be), etc., the angles included by 
the rays a and b, b and c, etc., so that also here (ab) = — (ba); 
(ab)+(bc)+(ca)=o. 

In analogy with the anharmonic ratio of four points, the anhar- 
monic ratio of these rays is 

sin (ac) /sin (ad) 
sin (be) I sin (bd) 

and may be designated by (abed) = k. 

What has been said about the permutations of four points 
applies without alteration to four rays. Consider now four con- 
current rays a, b, c, d passing through four points A, B, C, D of 
a straight line, respectively. From Fig. 7 it is seen that 

sin (ac) /sin (ad) CN /DM 



sin (ac) /sin (ad) 
(l) sin (be)/ sin (bd) = ' 



/ sin (ad) CJy / 
/sin (bd)~~CP/ 



sin (be)/ sin (bd) CP I DO' 

DM and CN being _L a and DO and CP _L to b. But 

CN AC J CP BC , 

DM = AD and DO = BD ] henCC 

CN /DM AC /AD 
CP / DO~ Bel BD 

(2) (abed) = (A BCD). 



i6 



PROJECTIVE GEOMETRY. 



This important result may be stated by the theorem: 

The anharmonic ratio of any four concurrent rays is equal to 

the anharmonic ratio oj four points formed by the intersection o) 

any transversal with these rays. (Pappus.) 




Fig. 7. 



In other words, if the rays a, b, c, d are considered as project- 
ing rays in a central projection, such a projection does not change 
the anharmonic ratio of four points. 

A system of rays in a plane and passing through the same 
point is called a pencil of rays. 1 By the above theorem all proper- 
ties of projective ranges of points may be transferred to pro- 
jective pencils of rays. 

In order to obtain an analytic expression for the rays of two 
projective pencils with the same vertices, assume the line repre- 
senting a projective range of points as the X-axis and the origin 
of the range as the origin of a Cartesian system. Let F, with 
the coordinates m and w, be the center of a pencil, then the equa- 
tions of the rays passing through the double points A and B of the 
transformation 



(3) 



x = 



(a— bk)y— ab(i — k) 
(i-k)y-(b-ak) 



(eq- 5> § 2) 



1 Cremona, loc. cit, p. 22. In the translation of Reye's Geometrie der Lage 
the term "sheaf of ravs" is used, while in Cremona's treatise "sheaf of rays" 
or "planes" means all rays or planes passing through a point in space. Ger. 
Strahlenbilschel. Fr. Faisceaux. 



PENCILS AND RANGES. tj 

are 

(4) nx-\- (a— m)y— an = o, 

(5) nx-\-(b— m)y— bn = o. 

Multiplying (5) by X and subtracting from (4), the equation 
of a third ray through V results: 

a— Xb \ a— Xb 



(6) nx-\- ( -. — mjy J n== °' 



This ray intersects the X-axis in a point, say D, whose abscissa 
d = 7-. To find the corresponding point C in transformation 

(3), put y = d= _ ; in (3). This gives for the abscissa c of C 

a—Xbk 
the value c = jt-, so that the equation of the ray passing 

through C becomes 
. . [a—Xbk \ a—Xbk 

(7) ^+(7-71— m )y-^-n; n=z °- 



i-xk r i-xk 



Comparing equations (6) and (7) with those of (4) and (5), 
we find, if (4) and (5) are written u = o, v = o, that (6) and (7), 
the equations of the rays VD and VC, are 

(8) u— Xv =0, 

(9) u—Xkv = o. 

For each value of X these equations represent a corresponding 
pair in a projective transformation of rays which is characterized 
by the anharmonic ratio k. In other words, for a variable A, 
(8) and (9) represent two projective pencils. 



18 PROJECTIVE GEOMETRY. 

§ 5. Involutoric Pencils. 

In the case of involution the anharmonic ratio is k= — 1, so 
that equations (8) and (9) of the previous paragraph become 

(1) u— Xv = o t 

(2) u+Xv=o; 

i.e., if u and v are any two rays, the rays u—Xv = o and u+Xv = o 
are harmonic with regard to u and v. For ^=0 and ^=00 the 
double-rays w = o and v=o of the involution are obtained. (1) 
and (2) define an involution of rays when X varies from — 00 to 

+ co . Suppose 

u = ax -\-by -\-c =0, 

v^a^+b^+c^o 

be the equations of the double-rays, so that (1) and (2) assume 
the form 

(3) (a- Xa l )x+ (b- Xbjy+c- ^ = 0, 

(4) (a+ Xa t )x+ (b+Xb 1 )y+c+ Xc t =o. 

The trigonometric tangents of the angles of inclination with 
+ X, or the slopes of (3) and (4), are 



(5) w = - 



a—Xa 



b-Xb, 



a-\- Xa t 

hence the tangent of the angle <j> included by (1) and (2) or (3) 
and (4) is 

m—m x 2(a l b—ab l )X 

(7) tan <f> - I + w . Wi - a 2+h2-x*(a*+b*)' 



PENCILS AND RANGES. 19 

This shows again that for A^=o and X = 00, tan <£=o, or <£ = o 
(180 ). In these cases the rays (3) and (4) coincide and the 
double-rays of the involution are obtained. Supposing that 
a 1 b—ab 1 ^o, which generally will be the case, we may ask for 

those values of A which will make tan </> = 00 , or (£>=-,& right 

angle. From (7) we find for this condition 

a 2 +b 2 -X 2 {a 2 +b 2 )=o y or 



a 2 +b 2 

which is always a real quantity. Whether we take the + or — 
sign for X in (8), we obtain the same couple of equations (3) and 
(4) ; hence the theorem : 

An involution oj rays always contains one, but only one, rect- 
angular pair. 

We shall now discuss the case where tan (f> = oo, or (56 = 90°, 
for all values of X. In order that this be the case, the quanti- 
ties a, b, a u b t must satisfy the conditions a 2 +b 2 = o, a 2 -\-b 2 = o, 
a 1 b—ab 1 ^o, or b = ±ia, b 1 =T ia u so that the equations 

ax-\-by-\-c = o, a 1 x+b J y-\-c t =o 

of the double-rays assume the forms 

c . c t 

x+iy-\--=o, x—ty+ — =o, 
a y a x 



and are imaginary. We can dispose of the constants - and — in 

such a manner that the double- rays will pass through the real 
point («,/?). Their equations then become 

^' \v = x— iy— (a — ify = o. 



20 PROJECTIVE GEOMETRY. 

The involution with these double-rays has only rectangular 
pairs. The equations of such a pair are 

\u-Xv = o, 

For real values of X the pairs are imaginary, since (10) may 
be written 



(ii) 



r-X 



n . .1— X , . , .i+^u . __. .„ . -. \ 

Putting ^^r~) = i", a real quantity, /= _ . . Thus, if in (10) we 

give X all imaginary values contained in the formula A — — , 

where [x is any real quantity, the corresponding pairs (n) in the 
involution will be real and rectangular. Now an involution of 
rays has generally only one rectangular pair and is determined 
by two pairs, hence the theorem: 

An involution o) rays having more than one rectangular pair 
has all its pairs rectangular. 

The double-rays of this involution are imaginary and pass 
through the two infinite points, which, as will be seen later on, 
are called the circular points at infinity, § 12. 

If an involution of rays shall contain the rays joining the 
vertex with the circular points, i.e., the two rays with the slopes 
+ i and —i as a pair, then according to (3) and (4) we must 
have 

a—Xa^ . a-\rXa x 

b^wr +tf b+xbr~ h 

or X(b x i—a x )-\-a—bi = o y 

X(b i t + a t ) + a+ bi = o. 



PENCILS AND RANGES. 



21 



These two equations must exist for the particular value of 
X which makes the slopes of (3) and (4) +i and — i. This can 
only be true under the condition 

(bji— a ± ) (a+ bi) — (b t i+ #1) (0— bi) = o, 



or 



aa 1 = — bb v 



Hence the theorem: 

Ij an involution oj rays contains the rays with the slopes -\-i 
and —i as a pair, then the double-rays oj this involution are per- 
pendicular to each other. 

Conversely, it can easily be proved that ij the double-rays oj 
an involution are perpendicular, then this involution contains 
the rays with the slopes -\-i and —i as a pair. The slopes of 
the rays of any pair in an involution, as defined by (3) and (4), 



a— Xa 1 a-\- Xa t 

^e -^ and ^b+W,' 
angles which these rays make with one of the double- rays, for 



Consequently the tangents of the 



instance u = o (slope 



are 



a a- 

b+b. 



Xa t 



Xb L 



X(ab 1 —afi) 



and 



1 + 



a a—Xa^ a 2 +b 2 —X(aa 1 +bb 1 ) 
b b-Xbi 



a-\-Xa 1 
T+Xb 1 



X(ab l —a 1 b) 



a a 
1 + b'b+Xb 1 



Xa ± a 2j r b 2j rX(aa 1 J r bb i y 



In case of perpendicular double-rays aa 1 +bb 1 = o J and these 
two tangents become equal. Hence the theorem: 

In case of perpendicular double-rays, the angles oj all pairs 
oj the involution are bisected by the double-rays. In such an 
involution two rays chosen jrom each oj two pairs include the same 
angle as the remaining two rays oj the two pairs. 



2 2 PROJECTIVE GEOMETRY. 



§ 6. Product of Projective Pencils and Ranges. 1 

i. In § 4 it has been shown that the equations of two pro- 
jective pencils of rays with the same vertex may always be writ- 
ten in the form 

(i) u— Xv = o, 

(2) u—Xkv = o. 

For every value of X these equations represent a correspond- 
ing pair of rays in a projective transformation which is charac- 
terized by the anharmonic ratio k. In other words, for a variable 
X (1) and (2) represent two projective pencils. Now the second 
pencil (2) may be moved into any other part of the plane with- 
out ceasing to be projective with regard to (1). This operation 
does evidently not change the general form of (2); only the 
expressions u and v are transformed into new expressions r and 
s. These represent two rays intersecting each other in the ver- 
tex of the moved pencil. Thus the equations of the two pencils 
are 

(3) u-Xv=o, 

(4) r—Xks=o. 

For each value of X there are two rays which intersect each 
other in a certain point P. If X successively assumes all values 
between — 00 and + 00 , P describes a locus whose equation is 
obtained by eliminating X between (3) and (4). This gives 

(5) vr—kus=o, 

1 The conception of pencils of curves and surfaces represented by equations 
of the form P+ XQ = o is due to Lame, who introduced it in his article, Sur les 
intersections des lignes et des surfaces, Gergonne's Annales, Vol. VII, 1816-17, pp. 
229-240. The generation of conies by projective pencils and ranges is due to 
Steiner (Systematische Entwickelung, etc.), who called it his "steam-engine" 
(Dampfmaschine) . 



PENCILS AND RANGES. 23 

an equation of the second degree in x and y, since u, v, r, s are 
linear in x and y. Hence the theorem: 

The product of two projective pencils of rays with separate 
vertices is a curve of the second order. 

2. Each linear expression, like ax+by-\-c = o, depends on 
two independent coefficients, so that the equation vr—kus = o 
contains eight independent coefficients. Arranging in (5) the terms 
according to powers of x and v, an equation of the form 

(6) ax 2 -\-2bxy-\-cy 2 +2dx-\-2ey J rf = o 

is obtained, where a, b, c, d, e, f are expressed in terms of the 
coefficients of u = o, v = o, r = o, s = o. 

As (6) contains only five independent coefficients, it is clear 
that the eight coefficients in w = o, v = o, r=o, s=o may always 
be selected in such a manner that (5) becomes identical with 
any equation of the form (6). See problem 11 in § 7. 

Hence the theorem: 

Every curve of the second order may be considered as the prod- 
uct of two projective pencils of rays. 



Fig. 8. 



(5) is satisfied by u = v = o and r=s=o, also by u = r=o 
and by v = s = o; i.e., the curve of the second order passes 
through the vertices of the projective pencils and also through 



24 PROJECTIVE GEOMETRY. 

the points of intersection of the rays v, s and u, r, Fig. 8. If, 

therefore, we want to write (6) in the form (5), we have to choose 

two points S and T on the given curve (6). Suppose that u = o 

and ^ = contain 5, and r=o and s=o T, then u, v, r, s, each 

only depend on one coefficient (slope), so that (5) depends on 

these four coefficients and k, which makes five coefficients in 

all. These coefficients may therefore be uniquely determined 

so that (5) represents or is identical with (6). Now two points 

00 (— i)oo 
S and T may be chosen in = go 3 different rays on a 

curve. Each of these determines a different but unique form 
of (5). The previous theorem may therefore be stated as fol- 
lows: 

Every curve 0) the second order may be produced in a doubly 
infinite number 0) ways by two projective pencils. 

At the same time we have proved the theorem: 

// two fixed points S and T be joined to a point P which 
describes a curve 0} the second order through S and T, the pencils 
(SP) and (TP) about S and T as vertices are projective. 

3. As the general equation of a curve of the second degree 
depends upon five independent constants, it is clear that five 
points of the curve determine it. Designating the coordinates 
of one of these points by x^ y { , there is 

axi 2 + 2bx i y i +cy i 2j r 2dxi+ 2ev;+/ = o, 

2* = !, 2, 3, 4, 5. 

These are five equations with five unknown quantities 

a 2b c 2d 2e n . 1 . , . . _ 

-7-, -J, -j, -7-, -j , which may be found by the usual method. 

Hence the theorem: 

A curve 0) the second order is determined by five points. 

In a similar manner it may be proved that two projective 
ranges of points can be represented by the equations 

(7) a-,«/? = o, 

(S) ?--/"^ = o, 



PENCILS AND RANGES. 25 

where a, /?, 7-, d are the line-equations x of four points in a plane. 
Assuming the knowledge of line-coordinates, the proof may 
be made without difficulty and may be left to the reader. 

For every value of pt there are two points which determine 
a straight line. If fi successively assumes all real values, this 
line envelops a curve whose equation is obtained by eliminating 
fx from (7) and (8) and which is 

(9) @y— Kad = o. 

This is an equation of the second degree in line-coordinates 
and consequently represents a curve of the second class. 2 In 
analogy With the previous statements we have also the theorems : 

Every curve of the second class may be produced in a doubly 
infinite number of ways by two projective ranges. 

If two fixed tangents S and T be intersected by a line P 
which envelops a curve of the second-class tangent to S and T, 
the ranges (SP) and (TP) on S and T as bases are projective. 

A curve of the second class is determined by five tangents. 



§ 7. Exercises and Problems. 

1. Assuming (ABCD)=k, find the values of the other 
twenty-three ratios which may be formed with the four points 
ABCD. 

2. Do the same when (ABCD) = + ij — 1. 

3. If X v X 2 , X 3 , X 4 and F 1? F 2 , F 3 , F 4 are corresponding 
points in a projective transformation, verify the relation 

*'."' ^,X 2 X 3 Z 4 ) = (F 1 F 2 F 3 F 4 ) by using 

ax+b 

y= r. 

J cx-{-d 

1 Line-coordinates of a line are the negative reciprocal intercepts of this line 
with the coordinate axes. The reader is referred to Salmon-Fiedler: Analytische 
Geometrie der Kegelschnitte, 6. ed., Vol. I, pp. 120-128. 

2 This statement stands for a definition. 



26 PROJECTIVE GEOMETRY. 

4. If the double-points of an involution are a = o and b=<x>, 
prove that the involutoric transformation has the form x+y = c. 

5. If x and y are a pair in an involution with the double- 
points a and b f prove the relation 

(x— a)(y— x) + (x— b)(y— x) + 2(x— a)(x— b) =0. 

6. Also establish the relation 



1 
x— a 



7. Prove that the middle point of an involution is always 
real. 

8. What is the form of an involutoric transformation if the 
double-points are -\-a and —a? 

9. An involutoric transformation referred to its center as 
an origin may be represented by x-y = k 2 , where ±k locates the 
double-points, (k may be real or imaginary.) Prove that the 
anharmonic ratio of the points represented by x, y,-\-k,— k is — 1. 

10. Prove that the rectangular pair of an involution of rays 
bisect the angles formed by the double-rays. 

11. The equation of a circle 

x 2 -\-y 2 = R 2 

is given, and on it the points (+r, o) and (— r, o). Find the 
equivalent equation 

vr— kus = o, 

where the pencils w— Xv = o and r—Xks = o have the given points 
as vertices. 

§ 8. The Complete Quadrilateral. 1 

In § 5 it has been found that if p and q are linear expressions 
in x and y, 

1 See Elemente der analytischen Geometrie by F. Joachimsthal, pp. 131-142. 



PENCILS AND RANGES. 27 

p=o, 
q=o, 

ap+Pq = o) (P \ 
ap-Pq = oy \<x /' 

always are the equations of four harmonic rays of a pencil. By 
means of this theorem it is now possible to study the harmonic 
properties of the complete quadrilateral. Let p = o, q = o and 




Fig. 9. 

r = o, 5 = be the equations of two pairs of rays, Fig. 9. The 
equations of the rays passing through the vertices of these pairs 
are cf the form ap-\-(3q = o and a'r+/? / s = o respectively. For 
certain values of a, ft and a', /?' these equations may both rep- 
resent a ray passing through the vertices of both pairs, so that 
we have the identity 

(1) ap+(3q=a'r+(3's. 
From this the further identities 

(2) ap-(3's=a'r-(3q, 

(3) p's- i 8q=ap-a'r 

follow. Identity (2) represents a straight line through the points 
of intersection of the rays p==o and s = o and of the rays r = o and 
q=o. The second represents a line passing through the points 
of intersection of p = o and r = o and of s = o and q = o. Adding 



28 PROJECTIVE GEOMETRY. 

(2) and (3) we get 

(4) ap-Pq=ap-Pq, 

i.e., the equation of a ray passing through O and P. The form 
of the equation shows that the ray is harmonic to the ray PQ 
with regard to the rays PC and PD. 

Identity (3) may be written ap—a'r^fi's—ftq. Subtracting 
this from (2), there results the new identity 

(5) a'r—fi f s=a f r—p f s. 

This is the equation of a ray through O and Q. The form of 
the equation shows that the ray QO is harmonic to the ray QP 
with respect to the rays QB and QC. As (4) and (5) result from 
(2) and (3) by addition and subtraction, it is proved that OP, OQ 
and AC, BD are harmonic pairs. PC, PD; QB, QC , BD, AC 
are called the sides, and OP, OQ, QP the diagonals, of the com- 
plete quadrilateral. The previous results may be summed up in 
the theorem: 

In every complete quadrilateral a pair of sides always jorms a 
harmonic pencil with the two concurrent diagonals. 

From this it follows that two vertices, for instance C and D, 
are harmonically divided by the two diagonals PO and PQ. 

Ex. 1. If p = o, q = o, r = o are the equations of the sides of a 
triangle, prove that any line of its plane may be represented by 
an equation of the form ap-\-ftq-\rT r = ' 

Ex. 2. Let p = o, q = o, r=o be the equations of the diagonals 
of the quadrilateral, prove that 

ap+ftq+irr^o, 

— ap-\-^q J {-j'r = o, 
OLp—@q+rr = o, 
ap J r^q—jr = o 

are the equations of a quadrilateral having those diagonals. 



PENCILS AND RANGES. 29 



§ 9. Perspective Pencils and Ranges. 

In § 6 it has been found that the equations of two projective 
pencils of rays with the vertices S and T may be written in the 
form 

K) \r + tis=o,* 

where u and v are two rays through S, and r and s two rays 
through T. In general the product of these pencils is a curve of 
the second order with the equation 

(2) us—rv = o. 

Every value of pt gives two corresponding rays u+ t uv = o and 
r+jis = o, which intersect each other in a certain point of the 
curve. We will now assume that the rays u, v through S and r, s 
through T are chosen in such a manner that there exists a value 
k of jj. so that the two corresponding rays u+kv = o, r+ks = o 
are identical, or that the ray through the vertices S and T is selj- 
corres ponding. In this case 

(3) u J r kv = r+ks. 
Eliminating u between (3) and (2) gives 

rs+ks 2 — ksv— rv = o, or 

(4) (r+ks)(s-v)=o. 

Eliminating v between (3) and (2) gives 

1 _ 1 
us— rs— t? +Tur = o. or 
k k 

(5) \s+yj(u-r)=o. 

1 Here ji = — \ and 5 ==ks in formulas (3) and (4), § 6. 



30 PROJECTIVE GEOMETRY. 

Equations (3), (4), and (5) show that in this case the curve 
of the second order degenerates into two straight lines, one pass- 
ing through S and T, the other passing through the points of 
intersection of u = o and r = o and of v = o and s==o. Hence the 
theorem : 

// the ray connecting the vertices of two projective pencils is 
self-corresponding, then the product oj the two pencils consists 0) 
the self-corresponding ray and another straight line. 

Two pencils of this kind are said to be in a perspective position, 
or simply in perspective. 

Similar arguments in line-coordinates, which may be left as 
an exercise to the reader, lead without difficulty to the theorem: 

If the point of intersection of two projective ranges is self- 
corresponding in both ranges, then the product {envelope) of these 
ranges consists of the self -corresponding point and another point. 

Two ranges of this kind are said to be in a perspective position, 
or simply in perspective. 

The line where corresponding rays of two perspective pen- 
cils meet is called axis of perspective. The point through which 
rays joining corresponding points of two projective ranges pass 
is called center of perspective. 

Ex. Prove the proposition concerning perspective ranges 
of points analytically (line-coordinates) and geometrically. 

§ 10. General Construction of Projective Pencils and Ranges. 

In § 2, 4 it has been proved that a projective transformation 
is determined by three corresponding pairs. This applies to 
pencils as well as ranges. This fact and the results of the pre- 
vious section make it possible to construct projective pencils 
and rays. 

A. Projective Pencils. — Let a, b, c and a' , b f , c f be three 
pairs of corresponding rays through the vertices L and I! re- 
spectively, Fig. 10. These determine two projective pencils of 
rays through the points L and If. Taking c and c' as bearers 
of two ranges of points, obtained by the intersections of a! , b' ', 



PENCILS AND RANGES. 



3* 



c f , . . . iand a, h, c with c and <?, respectively, we have accord- 
ingly the projective ranges 

{c-a'b'c' ...) = (c'-abc...). 

As the points (cc f ) and (c'c) are identical, it follows that they 
are in perspective, i.e., the lines joining the points {caf) and (c'a), 
(cb f ) and (c'fr), . . . are all concurrent, say at P. 




Fig. 



Hence, if any ray x of the first pencil is given, we know that 
the corresponding ray ocf will be situated in such a manner that 
the line joining (xc f ) and (x f c) will pass through P, and x f is 
found by joining the point of intersection of x and c' to P by 
a line £. The line joining U to the point of intersection of $ 



32 PROJECTIVE GEOMETRY. 

and c is the required ray xl. In an entirely similar manner 
any ray of the second pencil may be assumed and the corre- 
sponding ray in the second pencil be constructed. Any ray r t 
through P intersecting c and d in two points Y and Y' gives 
rise to two corresponding rays LY and L'Y', or y and /. From 
this construction it is seen that two projective pencils always 
admit of a third pencil which is in perspective with each oj them. 

Now it is known that two projective pencils produce a curve 
of the second order in a unique manner. The six rays a, b, c\ 
a', b r , c' determine the five points L, L', {oaf), (bb f ), (cd) of the 
curve, and every new pair of the construction like x and od, y 
and y } etc., determines a new point of the curve. The forego- 
ing construction gives us therefore a means to construct any 
number of points of a curve of the second order, as soon as five 
of its points are given. If L, L', A, B, C — in any order — are 
the given five points, join L and U each to^4, B, C, thus ob- 
taining the projective rays a, b, c and a', b f , c' '; then apply 
the construction and find as many points of the curve as 
desired. 

The ray / joining L and V is common to both pencils, but 
is not self-corresponding. Suppose t belongs to the pencil at L. 
To find its corresponding ray at Z/, produce t to its point of inter- 
section V with c' '; join T' with P and find the point of inter- 
section T of this line with c. The line joining U with T is the 
required ray t' . Following this construction in Fig. 10 it is clear 
that II T is nothing else than PL! . Similarly, if t f is considered 
as belonging to the pencil at L f , its corresponding ray will be 
PL. Taking a ray through either L or L', very close to /, and 
making the construction for the corresponding ray, supposing 
at the same time that the original ray passes to the limiting posi- 
tion of /, it is easily found that PL and P'LI are the tangents 
from P to the curve of the second order. 

B. Projective Ranges. — Let A, B, C and A', B', C be three 
pairs of corresponding points on the lines / and /' respectively, 
Fig. ii. These determine two projective ranges of points on 
I and /'. Taking c and d as vertices of pencils of rays, joining 



PENCILS AND RANGES. 



33 



A' , B f , C, . . . , and A, B, C, . . . , respectively, we obtain the 
projective pencils 

(C-A'B'C . . .) = (C -ABC . . .) 




Fig. ii. 



As the ray CO or C'C is common to both, it follows that they 
are in perspective; i.e., the points of intersection of the rays CA' 
and CA, CB f and C'B, etc., are all on the same straight line, 
say p. 

Hence, if any point X of the first range is given, the corre- 
sponding point X r is found by joining C to X and finding the point 
of intersection of this joining-line with p. The line joining C 
to this latter point cuts V in the required point X '. In an 



34 PROJECTIVE GEOMETRY. 

entirely similar manner any point of the second range may be 
assumed and the corresponding point in the first be constructed. 
Any point in p gives rise to two corresponding points on / and V . 
From this construction it is seen that two projective ranges always 
admit oj a third range which is in perspective with each oj them. 
The line p intersects / and /' each in a point whose corre- 
sponding points coincide with the point of intersection of / and /'. 
Again, I, I', AA', BB' ', and CC are five tangents to a curve 
of the second class and the foregoing construction makes it pos- 
sible — by joining X and X' — to construct any number of tan- 
gents. The line of perspective cuts / and /' in their points of 
tangency. 



§ ii. Exercises and Problems. 

i. Given five points of a curve of the second order; construct 
five other points, each being situated between two of the given 
points, i.e., one between A and B, one between B and C, etc. 

2. Construct the tangents at each of the given points. 

3. Given five tangents of a curve of the second class; con- 
struct any number of other tangents and the points of tangency 
of the given tangents. 

4. Two projective ranges (ABC . . .) = (A'B'C . . .) on the 
lines / and /' determine a curve K of the second class having 
AA', BB', CC, ... as tangents. Conversely, every tangent 
x of K cuts / and /' in two corresponding points of the ranges. 
If we now turn I' about its point of intersection P with / through 
the space containing K, Fig. 11, two coincident projective ranges 
arise. To obtain the double points of these, § 2, draw the 
bisector q of the angle between / and /'. The two tangents d l 
and d 2 , perpendicular to q, intersect either / or I' in the required 
double- or self-corresponding points D x and D 2 . x 

1 This construction has been successfully used as a base for the synthetic 
treatment of the projective continuous groups by Professor Newson and myself. 
See Kansas University Quarterly, Vol. IV, p. 243 and Vol. V, No. 1. 



PENCILS AND RANGES. 35 

5. What position must K have with respect to / and /' in order 
to make the projective ranges on / and /' involutoric? 

6. Show that with K as a circle the projective ranges are 
involutoric. 

7. Assume five points L, U, A, B,C of a curve of the second 
order in such a manner that the respective pencils are involutoric. 

8. Verify problems 1 and 2 on a given circle. 

9. Prove Newton's theorem (Principia, lib. i., lemma xxi). 
If two angles AOS and AO'S of given magnitude turn about 

their respective vertices O and O' in such a way that the point 
of intersection 5 of one pair of arms lies always on a fixed straight 
line u, the point of intersection of the other pair of arms will 
describe a conic (Cremona's statement). 

10. Prove Maciaurin's theorem (Phil. Trans, of the Royal 
Society of London for 1735.). 

If a triangle C'PQ move in such a way that its sides PQ, 
QC f , C'P turn round three fixed points R, A, B, respectively, 
while two of its vertices P, Q slide along two fixed straight lines 
CB f , CA f , respectively, then the remaining vertex C will describe 
a conic which passes through the following five points, viz., 
the two given points A and B, the point of intersection B f of 
the straight lines AR and CB f , and the point of intersection A' 
of the straight lines BR and CA f . 

§ 12. Projective Properties of the Circle. 

To specialize the results concerning projective pencils for 
the circle it is simplest to depart from the equation of the circle 

(1) x 2 -\-y 2 — r 2 = o. 
This may be written in the form 

(x+ iy) {x— iy) — r 2 = o, 

which itself may be considered as the result of the elimination 
of A between the projective pencils 

{x+iy+Ar = o, 

(2) 1 
[r+X(x-iy)=o. 



36 



PROJECTIVE GEOMETRY. 



The vertices of these imaginary pencils are the points of inter- 
section of the line at infinity, r = o, with the rays x+iy = o and 
x—iy = o. These points are called the circular points at infinity. 1 
Taking the center of the circle at (a, b), the equation of the 
circle becomes 



(x-ay+(y-by 



or 



(3) 



! (x— a) + i(y— b) } { (x—a) — i(y— b)} — r* 



o. 



Eq. (3) is the result of the elimination between the projective 
pencils 

x+iy— (a-\-ib)-\- Xr = o, 
r+ X(x— iy— a+ ib) =0, 



(4) 



and shows that all circles of the plane pass through the same 
circular points. As a curve of the second order is determined 
by five points, a circle must be determined by three points, two 
fixed points (the circular points) being given in advance. 

A circle can also be produced by two projective pencils with 
real vertices. Graphically this proposition is evident. If, in 

Fig. 12, ST be a chord of a circle, 
all angles subtended by this chord 
are equal, i.e., ZASC= Z.ATC, 
ZBSC= ZBTC, etc. Hence 

the pencils at S and T are projective. 
Connecting any point in space with 
all points of the circle, a cone is 
obtained. Cutting this cone by any 
plane and passing planes through the vertex of this cone and 
the rays of the pencils through 5 and T, two new pencils of 
rays (aJbjC^ . . .) and (a/b/c^d/ . . .) are obtained on the inter- 
secting plane, which are again projective. Their product is 




Fig. 12. 



1 Introduced by Poncelet, loc. cit., p. 94. 



PENCILS AND RANGES. 37 

therefore a curve of the second order; in this case a conic. It will 
be seen later on that all curves of the second order are identical 
with all conies. 

To show how a circle may be described as an envelope, assume 
first the line-equation of a circle 

(5) « 2 +^ 2 =^, 

where u and v are the line- coordinates and r the radius. Equa- 
tion (5) is the product of the two projective ranges 



(6) 



. X 

u+tv+- = o, 

— + X (u—iv) =0. 
r v J 



The coordinates of the line at infinity are u = v=o; conse- 
quently the points u+iv = o and u—iv = o are situated on the line 

at infinity. — = is the equation of the origin. Thus the pro- 
jective ranges (6) are situated on the two imaginary straight lines 
joining the points u+iv = o and u—iv = o with the origin. These 
lines are tangent to the circle and they pass through the circular 
points at infinity. A translation does not change these results, 
so that the theorem may be stated : 

The tangents to a circle from its center pass through the circular 
points at infinity. 

In the case of real projective ranges producing a circle it is 
more convenient to assume the circle and to prove that it is the 
product of two projective ranges. Let in Fig. 13 ZOAA' = a, 
Z.OBB' =/?, ZAOB = <f>, and in a similar manner /.OA'A=a' ', 
ZOB'B=fi, AA'OB f = ^. There is 2a+2a'=n- T , 2/3+2/3' = 
7i— f, hence /?— a = a f — /?. But /?— a = <f) and a'—fi f = (f) , i hence 
<f> = 4> f . This is true for any two tangents to the circle, so that the 
pencils (O-ABCD...) and (O-A'B'C'D' . . .) are projective. 
From this follows that the ranges formed by the points of inter- 



38 



PROJECTIVE GEOMETRY. 



section of all tangents with two fixed tangents are equal. Con- 
versely, the product of these particular ranges is a circle, as might 
also be proved directly. The tangents and ranges of this exam- 
ple may again be connected to a point in space. Cutting this 
configuration by any plane in space, two projective ranges pro- 




Fig. 13. 

ducing a curve of the second class are obtained. It will be seen 
later on that all curves of the second class are identical with all 
conies or curves of the second order. 



§ 13. Polar Involution of the Circle. 

Through a given point A, Fig. 14, draw any ray intersecting 
a given circle in two points C and D. On this ray determine a 
point B in such a manner that the anharmonic ratio 

(ABCD) = -i, 



i.e., harmonic. If this operation is repeated for every ray pass- 
ing through A, the points B on all these rays will form a certain 
locus which is a straight line, and which is called the polar of 



POLAR INVOLUTION. 



39 



the point A with regard to the given circle. The point A is called 
the pole. To prove this assume 




as the equation of the circle and (a, o) as the coordinates of the 
point A . The special position of point and circle has no influence 
upon the generality of the result. The equation of any ray 
through A may be written 



m 



y=(a— x)m. 



Solving (i) and (2) it is found that the abscissae x 1 and x 2 of the 
points of intersection C and D of the ray with the circle are 



(3) 



Now 



or 



x,= 



r+ am 2 -\-wr 2 -\- 2arm 2 — a 2 m 2 
1 + m 2 



r-\-am 2 —\/r 2 -\-2arm 2 — a 



nV 



2 1 + m 2 

{AB'X.X,) --= (A BCD) = - 1, 



a— x t 
b—x, 



a—x 2 
b—Xo 



from which 



b = 



2a— (x x +x 2 ) 



40 PROJECTIVE GEOMETRY. 

Substituting the values for x 1 and x 2 in this expression, then 

a— r 

i.e., the abscissa of B is independent of m and is therefore a con- 
stant. The locus of the point B is consequently a straight line 
parallel to the v-axis, or perpendicular to the line joining the point 
A with the center oj the circle. 

If A is without the circle, there are rays which do not cut the 
circle, or for which the points of intersection are imaginary. This 
is the case when r 2 +2arm 2 —a 2 m 2 <o, or a 2 m 2 —2arm 2 >r 2 , or 



\m\ > 



and x 2 are conjugate-imaginary, so that 



V a 2 — 2ar 

x t +x 2 and x x x 2 are real quantities and consequently also b is a 

real quantity. Hence if C and D are imaginary B is still real, 

±r 
and (ABCD) = — i. If m = > — , the points C and D coin- 

wa 2 —2ar 

cide and the rays through A become tangent to the circle, which 

are real when A is outside (2r< a), and imaginary when A is 

inside (2r>a). Hence the theorem: 

The polar of a point with regard to a circle passes through the 
points of tangency (real or imaginary) from this point to the circle. 

In the case of a pole within the circle the equation of the 
polar becomes 

a 

— — i 
r 

The greatest value for a is in this case 2r, so that up to this 

limit i<i and b>a. The smallest value of b is for a = 2r, 

r 

i.e., b = 2r. For a<2r, we have therefore always b>2r; the 

polar does not intersect the circle. For a = 2r, & = 2r, the pole 

coincides with the polar, which in this case becomes a tangent; 

i.e., a tangent is the polar of its point of tangency and a point of 

tangency is the pole of the corresponding tangent. For the center 

of the circle a = r and b= oo , the polar is the line at infinity. For 



POLAR INVOLUTION. 4r 

the tangents from the center m= ±i(a=r), so that the equations 
according to (2) become oe-\-iy=r, x—iy=r. This shows again 
that the tangents from the center of a circle touch the circle at its 
circular points, a result obtained in the previous section. 

§ 14. Continuation of § 13. 

Taking any point, for instance B, on the polar of A, it is 
clear that the polar of B must pass through A, since A is har- 
monic to B with regard to C and D as the other pair. Thus 
the theorem: 

The polar of a point which is situated on the polar of another 
point passes through the latter point. Conversely, the pole of 
a straight line which passes through the pole 0) a second line is 
situated on the latter. 

From this it follows that the tangents at C and D intersect 
each other in a point of the polar of A . This point is evidently 




Fig. 15. 

the pole of the ray (A BCD), through A. Using the results of 
§ 8, concerning the complete quadrilateral, it is now easy to give 
a simple construction of the polar of a point, or of the pole of a 
straight line. Through A draw any two rays intersecting the 



42 



PROJECTIVE GEOMETRY. 



circle in the points C, D and E, F, Fig. 15. Connect C with F, 
and Z> with E, and find the point of intersection G of these con- 
necting lines. In the same manner find the point of intersection 
H of the fines connecting C with E, and D with F. The line 
through G and H is the required polar of A . The proof is imme- 
diate, for (ABCD) = (AEIF) = — i, which is the condition that 
GH be the polar of A . The polar of H must pass through A , 
and since (HGBI) = — 1, it follows that it also passes through G. 
Hence the polar of H is AG. The polar of G passes through A 
and iJ, hence AH is the polar of G. The triangle ^4Gi7 
possesses the important property that the polar 0} each oj its 




vertices is the opposite side in the triangle, and the pole oj each 
side is the opposite vertex oj this side. This triangle is called 
a s el j- polar triangle with regard to the circle. 

Consider now in Fig. 16 the pole P and its polar p inter- 
secting the circle in two points A and B. Through P draw 
any ray c intersecting p in C, and determine the pole C of the 



POLAR INVOLUTION. 



43 



ray c. Then (ABCC) = — i. Designating the tangents from 
P to the circle by a and b, and the ray PC, which is the polar 
of C, by c f , there is also (abed) = — i. For every ray through P a 
pair of poles and a pair of polars are obtained which are harmonic 
to A and B, and to a and 6, respectively. In this manner an involu- 
tion of coincident poles and polars arises. In the case of the figure 
A and B are the real double-points, a and b the real double- 
rays of the involution. It is noticed that in this hyperbolic invo- 
lution each pair is separated by the double-elements. Two pairs 
either exclude each other entirely, like CC r and DD f , or include 
each other entirely, like DD' and EE r . If P were within the 
circle, we should have an elliptic involution, where two pairs 
always overlap each other. As an interesting example of this 
kind, consider the right-angle involution of the circle, Fig. 17. 




Fig. 



17- 



The polar of the center is the line at infinity. To every diam- 
eter a as a polar corresponds a pole A which is the infinite point 
of the perpendicular diameter a'. Thus a and a f arc a pair of 
the polar involution about the center. In fact the rays of each 
pair are perpendicular to each other. To find the double-rays let 



■mx. 



y= x, 



44 PROJECTIVE GEOMETRY. 

be the equations, of any pair. For a double-ray these equations 

must be identical. This is only possible when m= — -, or m 2 

= — i , which gives as the only possibilities m 1 = i, m 2 = — i. The 
equations of the double -rays are therefore x-\-iy = o and x—iy = o. 
As they are the double-rays of a right-angle involution, the' para- 
doxical result is obtained that each oj these rays is perpendicular 
to itself. Geometrically this has no meaning. 

Ex. i. Construct a self-polar triangle having two poles 
within the circle. 

Ex. 2. Discuss the elliptic pole and polar involution and 
make the necessary constructions. 

Ex. 3. Explain the involutoric relation between an inscribed 
quadrilateral A BCD of a circle and the quadrilateral circum- 
scribed at A, B, C, D. 



CHAPTER II. 

COLLINEATION. 

§ 15. Central Projection. 1 

A central projection, or a perspective, is determined by the 
plane of projection (plane of the picture) and the center (eye). 
Assuming the plane of the paper as the plane of projection and 
any point in space as the center, it is possible to construct the 
perspective of any figure in space on this plane. 

The center can most easily be located by a circle in the 
plane of projection. The radius of this circle is the distance 
of the center from the plane, and the center of the circle is the 
orthographic projection of the center upon the plane of pro- 
jection. This circle has been introduced into geometry by Pro- 
fessor Fiedler of Zurich, who calls it distance-circle 2 (Distanz- 
kreis). In this section only the projections of figures in a plane 
will be considered and the geometrical laws involved in this 

1 Historic Note. — Desargues, whom Poncelet called the Monge of. his cen- 
tury, was the first to investigate the relation of central projection to the geometry of 
position; i.e., the purely projective properties of central projection (perspective), 
in his Methode universelle de mettre en perspective les objets donnes reellement 
(Paris, 1636). These principles are also contained in the CEuvres de Desargues 
reunies et analysees par Poudra, Paris, 1864, Vol. I. 

Brook Taylor's New Principles of Linear Perspective, London, 17 15 and 17 19, 
k and J. H. Lambert's work, Die freie Perspektive, oder Anweisung jeden perspekti- 
vischen Aufriss von freien Stucken und ohne Grundriss zu verjertigen, Zurich, 1759, 
II. part, 1774, contain also the fundamental principles of perspective. 

For further information see the introductory chapter of Wiener's Darstellende 
Geometrie, Leipzig, 1884-87, which contains a history of this science and a chapter 
on perspective in Vol. II; also Fiedler's Darstellende Geometrie, Vol. I. 

2 D. Geometrie, Vol. I, 1883, and Cyclographie, Chapter VIII. The method 
followed here is that of Fiedler. 

45 



46 PROJECTIVE GEOMETRY. 

projection explained. The plane of projection will be desig- 
nated by tz' , and the arbitrary plane, whose perspective will 
be made, by tz, Fig. 18. Let 5 be the line of intersection of tz 
and 7r'. To obtain the projection P' of any point P in tz, con- 
nect P with the center C and determine the point of intersec- 
tion P f of this connecting line with tz' . In a similar manner, 
the projection V of a line / (RS) in tz is obtained as the line of 
intersection of the plane, passing through C and /, with tz' . 
From this construction the following fundamental laws are 
immediately clear: 

To every point of tz corresponds a point o) tz\ and conversely, 
and both points lie on a ray through C. 

To every straight line of tz corresponds a straight line of tz' ', and 
conversely, and both lines meet in a point of s. 

To the line at infinity of tz corresponds a line q' of tz' which is 
parallel to s. Conversely, to the line r' at infinity of tz' corresponds 
a line r parallel to s. 

The plane tz is usually determined by its trace s in tz' and 
either of the lines r and q' '. If a straight line / in tz is given, inter- 
secting s in S, the corresponding line I' is determined by drawing 
a line through C parallel to / and marking its point of intersection 
Q' with q' V It is apparent that Q' is the projection of the infinite 
point of /, and the projection of / consequently passes through 5 
and Q' '. Another way is to produce / to its point of intersection 
R with r and to join C with R. The line through S parallel to 
CR is /'. From the figure it is seen that CRSQ' is a parallelogram 
and that 

PS:PR = P'S:CR. 

The planes through C parallel to n and n f form a space of a 
parallelepiped. Keeping rJ fixed, it is possible to turn the planes 
tz and the planes through C parallel to tz and tz' down into tz, 
without changing 5 and q' and the distances of C and r, C and q', 
S and r, and 5* and q' in these planes. 

After the motion there is still CR \\ and = Q'S, and SP'=SP'; 
consequently the distances PR and PS are not changed by the 



COLLINEATION. 



47 



motion. From this it follows that after the motion P r and the 
revolved position of P lie on a ray through the revolved position 
of C. The laws expressing the geometrical relation between the 
revolved and the projected figure are therefore the same as those 




Fig. 18. 

between the figure in space and its projection. After the rabatte- 
ment, Fig. 18 assumes the form of Fig. 19. 

In this figure / and V are the two corresponding lines which 
with 5 and SC form a pencil of four rays through S. As this 
pencil is intersected by the rays CP and CQ, we have 

(CLP'P) = (CMQ'Q). 



The value of (CMQ'Q) is 



CQ' 
MQ' 



CO 
NO 



k, say; i.e., entirely in- 



dependent of the position of /, /', and CP. Thus, drawing any 
ray through C, intersecting s in S, and constructing any two 
corresponding points P and P f (rotated position of a point in iz 
and its projection on n'), we have 

(CSPP')= const. 



48 



PROJECTIVE GEOMETRY. 



Keeping CS fixed and constructing all possible pairs (P, P'), 
two coincident projective series of points are obtained having C 
and S as double-points. The different cases of central projection 
may be classified according to the position of the center and to the 
value of the constant k of the projection. Before entering upon 




Fig. 19. 

these details it is important to establish the analytical relation 
between a pair P, P'. 

Ex. 1. In both figures 18 and 19 it is noticed that the distance 
between r and s is equal to the distance between C and q'. As- 
sume the elements of a perspective as in Fig. 19, and draw the 
perspective of a triangle (a) which does not cut r; also of a triangle 
(b) which cuts r. 

Ex. 2. Draw in a similar manner the perspective of a regular 
hexagon which is not cut by r. 

Ex. 3. Draw the perspective of a circle 

(a) which does not cut r\ also of a circle 

(b) which cuts r in two points; and of a circle 

(c) which touches r. 



COLLIN E ATI ON. 49 

Ex. 4. Draw the perspective of a system of concentric circles. 

Ex. 5. Find V when / is parallel to s. 

Ex. 6. Construct the perspective of a circle having its center 
at C, Fig. 19. 

Note. — In all these exercises the given figures are, of course, 
in the revolved position of n\ i.e., in tt 7 . 



§ 16. Analytical Representation of Central Projection. 

In Fig. 19 assume any two perpendicular lines through C as 
coordinate axes and designate the angle which the X-axis makes 
with CO by $, and its angle with CP by <$>. Designate the coor- 
dinates of P and P f respectively by x, y and x r , y' '. Now 

iCLPP^K or CP>= C -^, or CP'=<|§^. 



From this CP' = 



k-CL-(k-i)CP' 



Now CP=Vx 2 +y 2 , CL^ j-. — jr. or, since 

J cos {(p—cpy 

, s t r • r • r X C0S </> J SU1 ^ 

cos (<£-<£)= cos ^ cos 0+sin </> sin <ft = _— +-==:, 



cz= 



x cos ([>+y sin 
hence, by substitution in the above value for CP', 

CN-Vx 2 +y 2 



CP'^ 



(1 — &) cos <f>-x+(i—k) sin (/r-y+k-CN 



50 PROJECTIVE GEOMETRY. 

Now xf = CP r ■ cos $ ; / = CP f • sin <j> ; hence 
3/ 



(I) 



CN-x 



(i-k) cos ^-#+(i-&) sin (p-y+k-CN' 



j 



CN-y 



(i — k) cos </>-x+(i — k) sin (p-y+k-CN 



In these expressions there are three arbitrary parameters : CN, 
k, <p. Conversely, if the transformation 



(ii) 



x! 



y 



ax 



dx+ey+f 

ay 
dx+ey+f 



is given, it always represents a perspective. To prove this, it is 
sufficient to reduce (II) to the form (I). This can be done in 

one and only one way, by putting — = k, 

(i — &) cos </> d (i — &) sin </> e 
= ~a~' 



CN 

e 
and as a consequence -r = tan (p. 

}-a 



CN 



From this CN 



Equations (II) are the most general 



Ve 2 +d* 

representation of a perspective. The points (x, y) in n for which 
(x', y') in 7if become infinite are evidently situated in the line 
dx-\-ey+f = o. This is therefore the equation of the line r. For 
the line s we have x=xf, y = y' '; hence from the first equation of 

(ii) 

dx 2 + exy +}x = ax, 
or dx+ey+f— a = o, 

as the equation of s. 

Equations (II) may also be written in the form 

x{ dxf — a) + yeaf + jxf = o, 
xdy' + y (ey f —a) + )y' = o. 



COLLIN EATION. 5 1 

The condition that the values for x and y become infinite is 



=o, 



(dx'—a) ex! 
dy (ey—a) 
or explicitly 

dx / +ey , + a = o. 

This is therefore the equation of q f . 

In these calculations the coordinate-origin is C, so that 



Vd 2 + e 2 
is the distance of s from C, or CN. 

The distance of cf from C is , and that of r from C is 

* Vd 2 +e 2 

This naturally all agrees with Fig. 19 from which they 



Vd 2 +> 

were derived. It must be remarked that these formulas only 

hold when C is in finite regions. 

§ 17. Special Cases of Central Projection. 

A. Involution. — If in Fig. 18 the center C is situated in 
the bisecting plane of it and rJ and if n is subsequently turned 
in the direction of the space between n and rJ in which the 
bisector lies, then r will coincide with q', and after the rotation 
CO = —N0. C and q / , r, in this case, are on the same side of s; 

CO 

k = jj^ = — i. This perspective is called an involution, since 

in Fig. 19 to every line / of tt corresponds a line V of rJ . If V is 
considered as the rotated position of a line in tt, then its corre- 
sponding line in rJ when rotated will coincide with /. This fact 
immediately . appears from the construction and also from equa- 
tions (II), which in this case assume the form 



an) 



ax 

x f = - 1 , 

dx+ey—a 

y >= a y 

J dx+ey—a' 



52 



PROJECTIVE GEOMETRY. 



or 



dxx r — a(x+x f )-\-ex'y=o i 
eyy' — a(y + /) + dxy' = o. 



Now in every perspective x'y = xy', as is seen immediately by 
dividing both equations (II) ; hence these equations remain the 
same when x, y and x f , y' are interchanged. 

If the plane n r is turned in the opposite direction as in the 
involution, k= + i, and the revolved position of n f in this case 
is also obtained from that of an involution by a reflection on the 
s-axis. 

B. Similitude. — This case arises when n' || iz\ i.e., if s is 
infinitely distant; hence the equation of s must appear in the 
form f—a=o and d = o } e = o. Equations (II) now go over into 



(IV) 



x — , x, 



y -jy- 



CP' 



When k = — is positive, (CLP'P) > o and equals -~p~ = ^ 

P and P r are on the same side of C, and the center in space is on 

the same side of n r and n, Fig. 20a. 




Fig. 20a. 




If k is negative, P and P f are on different sides of C, and the 
center lies between n' and 7r. If k = — i f the center is in the 
middle of n r and n and the perspective becomes central symmetry , 
Fig. 206. 



COLLIN E ATI ON. 53 

C. Affinity. 1 — By this term we designate a perspective 
whose center is at an infinite distance. All rays through C are 
parallel and intersect the axis of collineation s at a constant 
angle. To prove this, draw through every projecting ray, inter- 
secting iz f and 7z in two points P and P', a plane parallel to some 
fixed plane, and intersecting n' and n in the lines P'Q and PQ; 
where Q is a point in s, Fig. 21a. For all planes of this kind 




Fig. 21a. 

P'Q 

the triangles PQP' are similar, i.e., ^r= const; furthermore, 

for ever} 7 plane the lines P'Q and PQ include constant angles 

with 5. Hence, after revolving iz down into rc', Fig. 21&, and 

connecting again P with P', ZPQL = const., £P'QL = const., 

P'Q 

~p?y = const.; hence ZQPL= const., ZP'LQ = const., and PP r 

remains parallel to some direction cutting 5 at a constant angle. 

Now to J*:*?™® - Jg.«t; 

sin (PLQ) ^ 

1 Introduced by Mobius in his Barycentrische Calcul, p. 150. 



54 



PROJECTIVE GEOMETRY. 



similarly PL = ^^gp =P'Q . const. ; 



hence 



PL 



const. 



P'L 



The same result might be found from Fig. 19, where -p— = k. 

Formulas (II), however, are not valid in this case. To establish 
the analytical relation between P', (V, /) and P, (x, y) assume 
now s as the x-axis and any perpendicular to it as the y-axis. 




Fig. 21&. 



Let the constant slope of PP f be m; then the equation of the 
ray through P is, when £ and tj designate current coordinates, 
7] — y = rn(£— x), and the distance of L from O, Fig. 21&, is obtained 

mx — y 
as X = . Now from the figure A— x f = — k(x— X); eliminat- 
ing X, there is found 

(V) 



x f = x y, 

/ = ky. 



These are the equations of affinity. Conversely, if a transforma- 
tion 



(VI) 






by 



COLLIN E ATI ON. 



55 



is given, it may always be represented in the form (V), by putting 
b = k and = — m. A characteristic property of this trans- 
formation is that closed curves are transformed into closed curves, 
so that the areas enclosed by the two have the constant ratio k 
(in V). To prove this assume any triangle ABC and the corre- 
sponding triangle A'B'C. Designate the points of intersection 




Fig. 



of AB and A'B', BC and B'C, CA and C'A' by C^A,, B lt respect- 
ively. Now in Fig. 22 

(1) aABC = a4CA+ aBAJO^ aCB x A„ 

(2) aA'B'C'=aA'C 1 B 1 + aB'A.C^ aC'B.A,; 

but as the corresponding triangles on the left side have equal 
bases and altitudes of constant ratio k, we have 

AA'CiB^k- AiCA; AB'A x C x = k- aBA£ x \ 
ACB^^k- aCB,A x . 



Substituting these values in (2) and dividing (1) by (2), there 



is 



& A' B'C' = k- & ABC. 



Q.E.D. 



As these triangles may be infinitesimal, it follows by limiting 
summations that the same property holds for any corresponding 



5 6 PROJECTIVE GEOMETRY. 

areas. If k is negative, it follows that the area of A'B'C is also 
negative. For k= — i 

(VII) .j* , = *~m :y ' , 

which represents as a special case of affinity oblique axial sym- 
metry. For m= oo this goes over into orthogonal axial symmetry. 
(x ; = x, y' = — y). 

When k= + i and m^o an identical transformation is 
obtained. But the case is also possible where k= + 1 and m = o; 
i.e., where the rays PP f or A A ' are parallel to s, Fig. 23. In 



this case can have any value, say —A, so that the equa- 
tions are now 

(viii) \%; +Xy ' 

The effect of this transformation is that every point is moved 
parallel to s, and the amount of the motion is proportional to 
the distance of the point from s. As in (VII), equal areas are 



COLLI NEATION. 57 

here transformed into equal areas. This transformation is 
called elation. 1 



§ 18. Exercises and Problems. 

i. Given a straight line with the equation 

px+qy+r=o. 

Find the equation of the perspective of this line, and discuss its 
position with respect to the original line, q f , C, and s. 

2. Let x f y' = )(x') be the equation of a curve and suppose 
that j(x f ) is not divisible by x* and remains finite for x r =o\ then 
for this value of x', y' becomes infinite; in other words, the 
curve approaches the v-axis asymptotically. 

Applying the transformation (II) to this equation, or making 
a perspective of this curve, its equation becomes . 

a 2 xy(dx+ ey+ j) n ~ 2 = <j>(x, y), 

where <f>(x, y) is a polynomial of x and y of the wth order, pro- 
vided f(x) is an integral algebraic function of x of the wth 
degree. For x f =o, y' = oo . The corresponding values of x and y 

are x=o and y= . Thus putting in the above equation x=o, 

the value of y is found to be . The infinite branch of the 

curve is therefore transformed into a finite branch. This is gen- 
erally true, as it follows directly from equations (II) . If x f = oo , 

/ . ■ 

y = oo, --} = m (finite), the corresponding point x, y is necessarily 

situated in the line r, whose equation is dx+ey-}-}=o. Now 

y y r 

— = —} z =ni\ hence y = mx and from dx+enix+f=o, the coor- 

/ m\ 

dinates x= — ~ ri , y= — 3- of the required point are found. 

d+eni y d+em ^ r 

1 Term used by S. Lie, loc. cit. 



58 PROJECTIVE GEOMETRY. 

3. Find the transformed equation of xy=i and discuss it. 

4. Find the transformed equation of the circle 

f 2 
.,.2 1 ..2 / 





•* t y 


d 2 +e 2 ' 










Transform 


perspectively 


the curve 


/ = ^. 


For 


y= 


= 00, 


. Now y ■■ 


= i + x+ —.+ 
2! 


X 3 

lit-- 


hence 




+ 


1 + 



5. 

y =00 . 

x' x f2 /y\ 

—:+ — + . . . and lim I — ) = 00 . By equations (II) , 
2 . 3 • #' = 00 \^ / 



a y = dx+ey + f 

dx+ey+j 



Now for 3/ = 00 , y = 00 , we have 

— = -7 = 00 ana 
x x y 



y Y A * 

— = — =00 and — =0; 



hence r = e * * becomes r=e°=i; 

-X . j j f 

d—+e+ — e + — 

y y y 

hence y = and x=o. 

J a—e 

Make the corresponding construction. 

6. A perspective does not change the degree or class of a curve. 
As the curve is supposed to be algebraic, we may represent it 
by the polynomial f(x, y)=o, which evidently does not change 
its degree when transformed. Show this directly. 

7. Prove analytically that the transformation 

x' = x-\-ay, 
y = by 

transforms the area of the triangle ABC (x lt y x \ x 2 , y 2 ; x 3 , y 3 ) 
into an area of the triangle A'B'O (#/, y(\ x 2 ', y 2 '; x 3 ', y 3 ), 
so that AA'B'C' = b- A ABC. Use determinants. 



COLLIX E ATI OX. 59 



§ 19. Collineation. 1 

In the last two sections it has been seen that every perspec- 
tive transformation transforms a straight line into another straight 
line (into a point if the line passes through the center). The 
question is now whether there are other transformations with this 
property. From analytical geometry it is known that transla- 
tion and rotation are transformations of this kind. 

A. Translation. — By this operation all points of a plane 
are moved parallel to a certain fixed direction by the same amount. 
The equations are 

' = x+a, 
y+b. 



m {£ 



The slope of the direction in which (x, y) is moved is 

y f -y _b 
x r —x a' 



and the amount \Za 2 +b 2 . 

B. Rotation. — If even' point (x, y) is turned through an 
angle c3 about the center (origin), the coordinates xf > y f of the 
rotated point may be expressed by 

(X) j.v'^vcosp-vsin^ 

( y=x sin <p — y cos 9. 

(IX) and (X) are the equations of ordinary motions in a 
plane. If to a point (x, y) we first apply a rotation (X) and 
then a translation x?'=x'+a i y" = y r -b, and finally writing 
again x? and / for x" and y", the result is 

ix J = x cos cp-y sin $ + a, 
\y' = x sin <f>+y cos 9S4- b. 

1 Called by Chasles, in his Geometrie Superieure, art. 99, homographie; homo- 
graphic means to be in collineation. The word collineation goes back to Mobius' 
Barycentrische Calcid. 



60 PROJECTIVE GEOMETRY. 

Such a transformation changes only the position and not 
the shape of a figure, nor its size. Another transformation which 
does not change the character of a straight line is 

C. Dilatation. — The equations for these are: 

and may physically be illustrated by stretching a piece of rubber 
first in the direction of the #-axis and then in the direction of 
the ^-axis. Equal distances along one of the axes are stretched 
by the same amount. If one of the coefficients a, /? is i, then 

(XI) represents an orthogonal affinity. Combining the dilatation 
a/=ax, y'=(3y with the affinity x'^xZ + a^', y" =$$' and then 
dropping one prime clear through, the result is 

x? =ax-\-a 1 {3y, 

Applying to this the rotation x" = xf cos <j>—y' sin <f>-\- a' , 
y ff = x f sin </>+/ cos <£+fr', and dropping one prime, the result is 

xf =a cos <j) - x-\- (aJU cos <j>— /?/? x sin <f>)y+ a', 
y =a sin </>•#+ («i/? sin <£+/?/?! cos $)?+&', 

representing thus a combination of a dilatation, an affinity, a 
translation, and a rotation. Conversely, every transformation 

\x'=ax+by+c, 

(XII) W-dx+ey+f, 

represents such a combination. To prove this, put a=a cos <j> } 
d = 2sin(£; i.e., tan = -r, a=va 2 +^ 2 . Further, 

a$ cos <£—/?/?! sin <f> = b, 
afl sin <£+/?& cos <j> = e, 



COLLI NEA TION. 6 1 

from which 

n b cos &+e sin 6 . , . , bd+ae 

ir cos 2 + sm 2 <£ r r Va 2 +d 2 

ed—ab 
pp x = e cos <f>—bsm<p 



Va 2 +d 2 

and finally a'=c, &*— /. 

The transformations making up (XII) aW /eai'tf the infinitely 
distant line unchanged and transform areas into proportional areas. 
The same is therefore also true of their combination. Such a 
transformation is called a Linear Transformation, or Linear 
Deformation. As all of its constituents are projective, a linear 
transformation is also projective, i.e., it transforms pencils and 
ranges into projective pencils and ranges. Perspective and linear 
transformations are two of the most important projective trans- 
formations. 

D. General Collineation. — A perspective contains three 
arbitrary parameters, which is apparent when numerators and 
denominators of equations (II) are divided by a. Applying to 
the point x, y a linear transformation 

x' = ax+by-\-c, 
y'=dx+ey+f, 

and then to the transformed point x f , y' the perspective 



»"= 



y 



d^' + ey' + fc 
the result is, after dropping one prime, 

ax+by+c 



(XIII') 



(d^+ e$)x+ {dj)+e x e)y+fl 

dx+ey+f 

(d x a+ e x d)x+ {d 1 b+e 1 e)y+ } t ' 



62 PROJECTIVE GEOMETRY. 

These equations are of the form 

f , a 1 x+b 1 y+c l 
. -a B x+b 3 y+c 3 > 
{Xill) f _ a 2 x+b$+c 2 

[ y a 3 x+b 3 y+c 3 

Conversely, every expression of the form (XIII) can be repre- 
sented by (XIIP). To show this, put a t = a, b 1 = b } c x — c\ a 2 = d y 
b 2 = e , c 2^]\ d l a+e 1 d = a 3 , d 1 b+e 1 e = b 3 , } 1 = a 3 . From 

d 1 a-{-e 1 d=<i aj 
d 1 b-{-e 1 e J =b 3J 

a 3 e—b 3 d a 3 b 2 —b 3 a 2 

we rind d t = rr = —r 1 — , 

ae— bd a 1 b 2 — b x a 2 

b 3 a— a 3 b b 3 a x — a 3 b t 



*i = 



ae— bd aj) 2 — b ± a 2 



By means of these formulas it is possible to resolve every trans- 
formation (XIII) into a perspective and into a linear transforma- 
tion. The principal property of this transformation is that it 
transforms every straight line and every point projectively into a 
straight line and a point. It is the general projective transfor- 
mation of the plane, or a collineation in the plane. 

Dividing numerators and denominators of (XIII) by c 3 , it is 
seen that a collineation generally depends upon eight parameters. 
To prove that this is the most general transformation which trans- 
forms straight lines into straight lines, assume that 



y= p(*,y) 



Q(x, y): 

,_ *(*, y) 

y S(x,y) 



COLLIN EA TION. 6 3 

be a transformation of this kind; then the equation of every 
straight line 

ax' + by' + c = o, 

where a, b, c may have any real values, must be transformed into 
a linear equation between x and y. Thus 

aP(x, y)-S(x, y) + bR(x, y)Q(x, y) + cQ(x, y)-S(x, y)=o 

must be linear for all real values of a, b, and c. This can only 
be true if P, Q, R, and 5 are themselves linear functions of x 
and y. 

§ 20. Geometrical Determination and Discussion of 
Collineation. 

1. The equations of collineation depend upon eight parameters; 
these, when known, determine a collineation. If, therefore, any 
four points, of which no three lie in a straight line, are given: 
A x {x x , y ± ); ■ A 2 (x 2 , y 2 ); A 3 (x 3 ,y 3 ); A 4 (x 4 , v 4 ), we can assume any 
other four points with the same property as corresponding points 
of a collineation. That this assumption is legitimate is seen 
from the eight equations which may be established between the 
coordinates of the given points A ly A 2 , A 3 , J. 4 and A/, A 2 , A 3 , A/ 
by formulas (XIII). The eight independent parameters are now 
the unknown quantities which from the eight equations of condi- 
tion may be extracted in a definite manner. Hence the theorem : 

There is one and only one collineation which transforms a quad- 
rilateral in a plane into any other quadrilateral of the same plane. 
Two quadrilaterals in a plane determine a collineation uniquely. 

2. An important problem in a collineation is the determina- 
tion of those elements, points, or straight lines which are not 
changed in position, i.e., of the invariant elements. To find the 
straight lines which are invariant, assume their equation in the 
form 

(1) ax' + by' + c = o. 



64 PROJECTIVE GEOMETRY. 

By the collineation (XIII) this is transformed into the equation 

(2) (##!+ &&! + cc^)x-\- (a# 2 + bb 2 4- cc 2 )y+ (aa 3 + bb 3 -\-cc 3 ) =0. 

This will be identical with (1) if the three equations hold 
aa 1 -{-bb 1 +cc 1 = Xa, etc., or 

f a(a 1 — X) + bb 1 +cc t =0, 

(3) <aa 2 + b(b 2 -X) + cc 2 =0, 
( aa 3 +bb 3 + c(c 3 — X)=o. 



These are consistent only if the determinant 

(4) 



a, — X b x c x 

&2 °2~~ A C 2 

a 3 b 3 c 2 — X 



o, 



which gives a cubic equation for the proportionality-factor X. 
Solving (4) for X and substituting any of its values in (3), we can 

easily extract the values of — and — from any two of equations (3). 

c c 

These values inserted in (1) give the equation of an invariant 
straight line of the collineation. As there are three values for X, 
there are three such lines. Hence the theorem: 

A collineation in a plane leaves a triangle invariant. 

The vertices of this triangle are invariant points, while other 
points of the sides of a triangle are generally transformed into 
other points of the same sides. (See ex. 6 in § 23.) It may 
happen that two roots of the determinant (4) are conjugate 
imaginary, so that the invariant triangle in this case has only 
one real side and one real point (above example). 

3. In equations (XIII) both x f and / become infinite for all 
points of the line 

a 3 x+b 3 y+c 3 = o] 

hence, in the collineation, to this line corresponds the line at 
infinity. Solving equations (XIII) for x and y, the result is 



(5) 



COLLIN EATION. 65 

_ (Vs- V2K + ( Vi- ^i g 3 )/ + ( v 2 - y 1) 

j ^ (a 3 c 2 - fl 2 c 3 )3/+ (g^ 3 - flapy + (a 2 Ci- ^2) 
[ ^ "" (o 2 &s- ^K + (^1- aA)/ + ( a A~ aj>i) ' 



From this it is seen that all points of the line at infinity, x = 00, 
y = oo are transformed into the line 

(6) (fl 2 & 3 - 03&1K + (a A— «A)/+ K& 2 - a A) =0. 

4. Suppose a collineation (XIII) has been applied to a plane. 
Turn the transformed plane through an angle about the origin 

and translate it afterwards in the direction tan # = — through a 



distance \ / a 2j r b 2 . The result of these successive transforma- 
tions of the original plane (x, y) is expressed by the equations 

(Oi cos 4>—a 2 sin <£+ aa 3 )x+ 
(b 1 cos <£ — b 2 sin <£ + afo,)}' + c t cos <jS — c 2 sin <jS + #c 3 
a^+b«y+c 3 ' 

(a x sin <£+a 2 cos ^>+6a 3 )x+ 
f/ (& x sin <£ + 6 2 cos <£ + bb 3 )y + q sin <£ — c 2 cos + fc 3 

The angle <f> , and a and 5, can always be determined in such a 
manner that 

b x cos <£— & 2 sin <f> + ab 3 = o, 
a x sin 0+a 2 cos <j)+ba 3 = o, 
a x cos 0— a 2 sin <fi+aa 3 = b l sin <£+ 6 2 cos <£+ bb 3 , 

so that by this motion (0, a, 6) the equations assume the form 

ax+ p 



x" = 



a^+b^+c^ 






66 



PROJECTIVE GEOMETRY. 



If the original plane is translated in such a manner that x = x x - 



a' 



y = y 1 , the connection between the (x" , /') plane and the 

( x i> yd plane will be of the form 



x" = 



ax, 



a^x+fcy+n' 

ay 1 



Hence the theorem: 

// two collinear planes (figures) are given, it is always possible, 
by proper motions, to bring both into a perspective position. 

And as a corollary: 

Any two quadrilaterals in a plane can always be moved into a 
perspective position; one may be considered as a perspective of the 
other. 

As a special case we have: 

Any quadrilateral may be considered as the perspective oj a 
square; and conversely. 

On account of its practical importance this proposition will 
be treated graphically in § 27. 



§21. Continuous Groups of Projective Transformations. 



If to the point x f , / obtained by the projective transforma- 



tion 



(1) 



a 1 x J r b i y J r c 1 
a s x+b 3 y+c 3 ' 
a 2 x+b 2 y+c 2 
a 3 x+b s y+c 3 

another transformation of the same kind 



as^+Ay+rs' 

tfs^+Ay+ra 



COLLIN EATION. 



67 



is applied, the result is 



(3) 



00 Asx+Bsy+Cs' 
A 2 x+B 2 y+C 2 

y = 



A 3 x+B 3 y+C 3 ' 
where 

A^afr+P^+rjOs, B 1 =a 1 b 1 -\-P 1 b 2 J rTx h ^ C^a^+^+y^ 
A 2 =a 2 a 1 -\-p 2 a 2 +y 2 a 3 , B 2 =a 2 b 1 +{3 2 b 2 +y 2 b 3 , C 2 =a 2 c x +fi 2 c 2 +Y 2 c 3 , 
A 3 =a 3 a 1 +p s a 2 +y 3 a 3 , B 3 =a 3 b 1 +(3 3 b 2 +r 3 b 3 , C 3 =a 3 c 1 +/? 3 c 2 +r 3 ^3- 

Transformation (3) which changes (x, y) directly into {x" ', /') 
is of the same form as (1) and belongs, therefore, to the totality 
of all projective transformations. For this reason it is said that 
all projective transformations of the general type form a group. 
It is eight-termed (achtgliederig), since its general equations 
depend upon eight parameters. In § 20 we saw that every trans- 
formation (1) leaves a triangle invariant, and this fact is the 
characteristic property of the general projective group. 

It is not our purpose to discuss all possible projective groups, 
and we shall simply point out the most important ones. 

The six-termed linear transformation 



(4) 



x f =a t x +b l y+c v 
y' = a 2 x+b 2 y+c 2 



is clearly a group. As it is contained in the general group it is 
called a six-termed subgroup of (1) and leaves the line at infinity 
invariant. 

The perspective 



(5) 



a 3 x+ b 3 y+ c 3 ' 



y= 



a^x+b 3 y+c 3 



68 . PROJECTIVE GEOMETRY. 

is a three-termed subgroup leaving a point (origin x=o, ^=o) 
and the axis s of collineation (every point of it) invariant. As in 
these examples, it may be found that every projective special 
group leaves a certain figure invariant. The particular invariant 
figure is characteristic for the group. 

The theory of continuous groups is a creation of Sophus Lie x 
and is of the greatest importance in various branches of mathe- 
matics, notably in the theory of differential equations. 



§ 22. The Principle of Duality. 2 

A. Two forms of projectivity have already been studied (§§6, 
9, io, 13) — the projectivity of pencils and that of rays. Two 
projective pencils generate a curve of the second order; two pro- 
jective ranges generate a curve of the second class. In the first 
case the point is the generating element of the curve; in the 
second it is the straight line. In both cases the equations in point- 
and line-coordinates are respectively of the second degree. 

A plane figure may therefore be considered either as a configura- 
tion of points or as a configuration of straight lines. This is the 
principle of duality. Two figures are called dual if to a point in 
one corresponds a straight line in the other, and conversely. 
Below is a scheme of some dual figures, which by the foregoing 
statements explains itself. 



1 Vorlesungen iiber continuierliche Gruppen. Theorie der Transformations- 
gruppen. 

The theory of projective groups has been worked out synthetically by Pro- 
fessor Newson and partly by the author himself. See Kansas University Quar- 
terly, Vol. V, No. 1. 

2 Historic Note. — Poncelet in his Traite, 1822, was the first geometer who 
showed by his method of reciprocal polars the great importance which dualistic 
relations have for geometry. Gergonne, in the Annales de Mathematiques, T. 
XVI, 1826, stated the principle of duality in all its generality and independently 
of reciprocal polars. Plucker first established the analytic expression for duality, 
and Steiner gave the equivalent geometric interpretation. 



i. Range of points on a straight 
line. 



2. Curve. 



COLLIN EATION. 69 

Pencils of rays through a point. 



3. Triangle. 

4. Quadrangle. 



•C 

A -B 




•B 



5. Points of a plane (00 2 ). 

6. Point of a curve. 

7. Tangents from a point to a 

curve. 

8. Point of intersection of two 

straight lines. 

9. Intersections of curves. 




Envelope. 



Trilateral. 



Quadrilateral. 



Straight lines of a plane (00 2 ). 

Tangent of a curve. 

Points of intersection of a straight 

line with a curve. 
Straight line connecting two 

points. 
Common tangents of curves. 




B. Analytically the principle of duality is expressed by the dis- 
tinction between point- and line- coordinates : x, y and u, v. If 
the point with the coordinates x and y satisfies the equation 

ax-\- by-\-c = o, 

then the point is situated on the straight line which cuts from the 

x- and ^-axes the distances and — 7-. If x and y are kept 

a b 

fixed and a, b, c, or u = — , v = — are varied, then for all values of 
c ' c 

a, b, c, or u and v, which satisfy ax+by+c, or ux+vy+i=o, a 

straight line is obtained which passes through the point (x, y). 

Hence 

ux+vy+ 1=0 

is the quation of the point (x, y) in line-coordinates w, v. In 
§ 19 the equations for a general collineation were established for 



70 PROJECTIVE GEOMETRY. 

point-coordinates. The problem now is to do the same thing 
for line-coordinates. A straight line 

(i) ax' + by' + c = o 

by equations (XIII), § 19, is transformed into 

x(aa ± -f- ba 2 + ca 3 ) + y (a b x + bb 2 + cb 3 ) + ac t -f bc 2 + cc 3 = o, 

or 

aa 1 +ba 2 + ca 3 ab^bb^cb^ _ 
^ ' ac x +bc 2 +cc z y<Lc x +bc 2 +cCz ~ 

Putting a^' + fry' + c into the form ux f -f- vy' + 1 = o and (2) 
into the form u'x+v f y+ 1 = 0, the required equations of collinea- 
tion in line-coordinates 



(XIV) 



~ C X U+C 2 V+C 3 

, bjU+bzV+bs 

v — , 

c x u-\-c 2 v+c z 



are obtained. By this transformation every straight line with the 
coordinates u, v is transformed into a straight line with the coor- 
dinates «', 1/. 

The discussion of these formulas is similar to that of (XIII) 
and may be left to the student. In (XIII) and (XIV) the ana- 
lytical expressions for the dualistic interpretation of collineation 
have been obtained. As for (XIII), the group-property is funda- 
mental for (XIV). In case of perspective in line- coordinates, a 2i 
a 3, h> b 3 vanish from (XIV). 

§ 23. Exercises and Problems. 

1. Show that the transformation 

x f = x cos <j)+ y sin </>, 
y = x sin 4>—y cos ^>, 



COLLIN EATION. 7 1 

consists of a rotation through an angle cf> and a reflection on the 
x-Sixis. 

2. What becomes of the circle x 2 +y 2 = r 2 ' after a dilatation; 
what is the ratio between the enclosed areas before and after 
dilatation ? 

3. Investigate the transformation 

yf = ax-\- by, 

y' = cx+dy, 
where ad—bc=i. 

4. The area included by a closed curve C in the a//-plane is 

obtained by evaluating / dx'dy' '. If we now transform the x'y'- 

plane by the equations x' = <j)(x, y), y=<l>(oc,y)i the area of the 
transformed curve c is 



-im-tw^'- 



A 

Take now a linear transformation 

x r — ax+by+c, 
y f = dx+ey+f, 

then the area A f of a curve in the transformed plane expressed 
in terms of the area A of the original curve is 

A' = (ae-bd)A, 

A f 
or -r = ae—bd: 

A ' 

i.e., in a linear transformation corresponding areas have a con- 
stant ratio. 

5. Prove that all points of the xv-plane are transformed into 
a straight line when ae—bd = o. 

1 Picard: Traite d' Analyse, Vol. I, pp. 98-102. 



72 



PROJECTIVE GEOMETRY. 



6. Find the invariant points from the equations of a general 
collineation. In these equations set x=— , y=— i xf =— n y' —-^-j 
and designate by X a factor of proportionality; then 



(1) 






If (£'> */> CO is identical with (6, >?, C)> we g et the condition 



(2) 



or 



(3) 






a x — ^ 


&! 


^1 


<2 2 


& 2 -^ 


c 2 


<*3 


^3 


c 3 -X 



o. 



This determinant gives three values for X, consequently from 
(2) three sets of values for (£, f), C). Two values of X, hence 
two of the invariant points, may be imaginary, while the line join- 
ing them is real (§20, 2). 

7. Show that all motions in a plane form a group. 

8. Prove the same for affinity; 

9. For symmetry (central and orthogonal); 

10. Similitude. 

11. Find the invariant elements of an affinity. 

12. How does a linear equation affect an hyperbola? 



§ 24. Orthographic Projection. 

1. In an orthographic projection two perpendicular planes 
of projection are assumed. One is in a horizontal position, 
and is designated by H; the other in a vertical position, and is 
designated by V. Both H and V intersect each other in the 



COLLIN EATION. 



73 



ground-line GL and divide the whole space into four quadrants, 
or angles, Fig. 24. With respect to the observer, space may be 
described as above or below H, in front or back of V. The four 
angles are now numbered as follows : 

I Angle : above H, in front of V. 

II Angle: above H, back of V. 

III Angle: below H, back of V. 

IV Angle : below H, in front of V. 




Fig. 24. 



In any of these angles, the projections of a point are obtained 
as the foot-points of the perpendiculars (projectors, projecting 
lines) from these points to H and V. If A is the point in space, 
we may designate the horizontal projection of A, which is in 
H, by A', the vertical projection of A by A" (in V). 

Let the plane through A A' and A A" intersect GL in A*, 
then there is A"A*=AA', A'A*=AA". In order to have the 
representation of these projections in one plane (plane of the 
drawing; blackboard), one of the planes, for example V, may be 
rotated about GL, so that the part of V above GL turns from 



74 



PROJECTIVE GEOMETRY. 



the observer till it coincides with H. After the rotation the 
upper portion of V covers the back part of H, and the lower 
portion of V lies in coincidence with the front part of H. Accord- 
ingly the projections of points in the different angles will lie as 
follows with respect to GL: 



iJ-projection below, F-projection above GL; 
above. above " : 



Point 


in 


I Angle 


a 


a 


II 


u 


u 


a 


III 


a 


l( 


a 


IV 


a 



above, 
below, 



below 
below 



The same is true of the projections of any figures situated in 
the different angles. 

In Fig. 25 these cases are represented. The two projections 
oj a point necessarily lie in the same perpendicular (eventually 



J c 

f 

G .]' i | L 


A 4 

D'6 



Fig. 25. 

extended) to GL. A fixed point in space can be represented in one 
and only one way by two projections. Conversely, two points in 
the same perpendicular to GL always represent the projections oj 
a point (but only one point). 

2. Straight Line. — A straight line is determined by two 
points (the only line or curve determined by two points). An 
orthographic projection (oj course, also a perspective) oj a straight 
line is also a straight line. Hence, if A' , A" \ B' ', B" are the 
projections of two points, the lines joining A' , B f and A" , B" 



COLLIN EA TION. 



75 



are the corresponding projections of the line AB. Generally 
any straight line — when produced — pierces H and V. The 
points where this occurs are called the traces of the line and may 

conveniently be designated by 



*i, t 2 



etc. (/^horizontal, 




Fig. 26. 



i 2 = vertical trace). A point P is 

situated on AB when P f is on 

A'B', P" on A"B", and 

P'P"LGL, Fig. 26. To find 

the traces t 1} t 2 when A r B r and 

A"B" are given we notice that 

the F-projection of a point in H 

lies in GL, and that the iJ-pro- 

jection of a point in V lies also 

in GL; the other projections coincide with the points themselves, 

respectively. 

Hence, to find the horizontal trace t ± of AB, produce A"B" 
till it intersects GL at t". This, being still situated on A"B" , is 
the vertical projection of a point of AB, and as it is also on GL, 
the point must be in H and is necessarily the required trace t v 
Hence, to find t l3 produce the perpendicular to GL at t x " till it 
intersects A'B' produced at the required t v Similarly, t 2 is found 
by drawing a perpendicular at the point of intersection t 2 of 
A'B' with GL and producing it till it meets A"B" produced at 
the required t 2 . 

It is evident that any two lines I ' and I " may he considered as 
the projections of a line I. This line is uniquely determined by 
l' and l' r . To prove this draw any two perpendiculars to GL, 
cutting / ' and / " at A ' , A " and B' , B" '. These points, however, rep- 
resent the projections of two points and hence the problem is 
reduced to the foregoing considerations. This proposition is 
altogether general no matter how the lines may be situated. 
// one line is perpendicular to GL, the other reduces to a point. 
Usually the projections may be assumed indefinitely extended. 

A finite portion may be cut out by two perpendiculars to 
GL. If the traces t x and t 2 are given, the projections are found 



76 PROJECTIVE GEOMETRY. 

by drawing the perpendiculars t£" and t 2 tl to the ground-line; 
then tj 2 ' is the iJ-projection, t 2 t" the F-projection of the Hne. 
3. Plane. — A plane is determined 

(a) By three points not in the same straight line; 

(b) one point and a straight line ; 

(c) two intersecting lines ; 
{d) two parallel lines. 

The most convenient manner to represent a plane is by its 
traces, i.e., its lines of intersection with the planes of projection. 
The traces of a plane meet in the ground-line and may be desig- 
nated by a v o 2 ; r lf t 2 ; etc. Let 5 and T be the points where 
<j x and o 2 , t t and t 2 meet. A plane in a general position extends 
into all four quadrants, and if nothing else is specified it will 
be understood that the plane is indefinitely extended. 

Ex. 1. Draw the projections of a straight line 

(a) _L H (/) _L GL 

(b) _L V (g) in a plane _L GL 

(c) || H (h) cutting GL 

(d) || V (i) iniJorF 

0) || GL (j) in V and _L H, 

and repeat the construction in all four angles; also construct the 
traces in every case. 

Ex. 2o Draw the traces of a plane 

(a) ± H (e) || H 

(b) ± V (/) || V 

(c) _L H and V (g) passing through GL 

(d) || GL (h) solve the foregoing problems 

in all four angles. 

Ex. 3. A profile plane is a plane _L GL. Given a plane |l GL; 
find its distance from GL. 

Ex. 4. Given any plane; locate a point in this plane; i.e., draw 
its projections. 

Ex. 5. A straight line lies in a plane when its traces lie in the 
corresponding traces of the plane; conversely, a plane passes through 



COLLIN EATION. 



77 



a line if its traces pass through the corresponding traces of the line. 
Construct the traces of the planes determined by (a), (b), (c), or 
{d) under 3, § 24. 



§ 25. Affinity between Horizontal and Vertical Projections of 
a Plane Figure. 

RABATTEMENT. 

i. The orthographic projections of a figure in a plane o 1 Sa 2 
are perspectives with infinitely distant centers in directions per- 
pendicular to H and V. In this case any of the projections and 
the corresponding revolved figure are in the relation of affinity 
(orthogonal affinity with one of the traces as an axis). But there 
exists also affinity between horizontal and vertical projections of a 
plane figure, as we shall now see. 




Fig. 27. 

2. Let P be the bisecting plane of the first and third angles, 
Q the bisecting plane of the second and fourth angles. The pro- 
jections 0) a point in P are equally distant from GL; those of a 
point in Q coincide. 

Hence, to find the point of intersection 0} any line I with Q, 
produce I ' and I " till they intersect at X; X represents the coinciding 



7« 



PROJECTIVE GEOMETRY. 



projections of the required point of intersection, Fig. 27. If this 
point X is infinitely distant, i.e., if /' || I" , then / is parallel to Q. 

If a plane P is given, then all lines in this plane will generally 
intersect the line of intersection s of P and Q. Hence the cor- 
responding projections of all lines in P will meet in points of 




s', 5" ; this line is therefore the axis of affinity which exists between 
the projections of figures in a plane P. If A is a point in P 
(traces s l9 s 2 ), then we can rabat P about s t into H. During 



COLLIN EATION. 



79 



the rabattement A describes a circle with center in s t ; A' describes 
a perpendicular to s v A line /, its horizontal projection /', and 
its rabatted position V" meet in the same point of s l9 Fig. 28. 
Hence there exists also affinity between the horizontal projec- 
tion of a plane figure and its rabattement into H. In Fig. 28 
these affinities are illustrated in case of a triangle. If the hori- 
zontal projection of a triangle, 
A'B'C and A" are given, B" 
and C" may be found by applying 
the principle of affinity. Thus 
A 'B' meets A"B" in a point r 
of 5 (Y, s"), and B', B" lie in a 
perpendicular to GL) hence join 
A" with r and through i?' draw 
B'B" ±GL, thus determining B" 
Similarly C" may be obtained. 
By the same principle the rabatte- 
ment A"'B["G"' may be con- 
structed if one point, say A f// , is 
known. An interesting special case 
is obtained when s 1 and s 2 are 
equally inclined towards GL. Then 
s is perpendicular to GL and corresponding projections of closed 
figures have equal areas, Fig. 29. This case has been discussed 
in § 17, Fig. 23, elation. 

Ex. 1. Discuss the affinity which results when s x coincides 
with s 2 . 

Ex. 2. What must be the position of a plane P to obtain 
orthogonal affinity, § 17? 

Ex. 3. What plane gives orthogonal symmetry? 

Ex. 4. What is the position of a plane if the projections of 
any point in this plane are always equally distant from the axis of 
affinity 5? (Oblique symmetry.) Make use of a profile-plane. 

Ex. 5. A straight line is perpendicular to a plane if its pro- 
jections are perpendicular to the corresponding traces 0} the plane. 
Prove this proposition. 




Fig. 29. 



8o PROJECTIVE GEOMETRY. 

Ex. 6. What is the position of a plane whose traces coincide ? 

§ 26. Homologous Triangles. 

1. In § 15, treating of central projection, two planes n and t! 
intersecting each other in s and a center of projection V (C) were 
assumed. Consider now any triangle ABC in n and find its 
projection A'B'C in n'\ then a pyramid with base A'B'C and 
vertex V is obtained which by the plane n is cut in the triangle 

ABC. Revolving n and with 
it ABC about s down into -', 
thus assuming the position 
A X B X C X , then from the laws of 
perspective we know that A'A lt 
B'B X , C'C X are concurrent at 
a point W, and the points of 
intersection of A'B' and A X B X , 
B'C and Bfi^ C'A' and C X A X 
are collinear, i.e., lie on the 
same straight line. Triangles 
with this property are called 
homologous, x Fig. 30. Any two 
triangles for which the lines 
joining corresponding vertices are concurrent may always be. 
considered as resulting by the foregoing projection and rabatte- 
ment, so that the following theorem holds: 

If two triangles AJ$ x C r anal A 2 B 2 C 2 are situated in such a 
manner that the lines joining corresponding points like A x A 2y 
B X B 2 , C X C 2 are concurrent at V 3 , then the points of intersection 




.-:®V S 



Fig. 30. 



A B ) 

of corresponding sides, in symbols . 1 „ x > y, 

are collinear on s. 



B,C 2 



(v, 



Zy 



1 Casey in his sequel to Euclid uses the word homologous. Mr. Leudesdorf 
in his translation of Cremona's Projective Geometry, p. 10, uses homological. See 



also Poncelet, loc. cit. 



COLLINEATION. 



8l 



Conversely : 

If two triangles are situated in such a manner that the points 
of intersection a, /?, y of corresponding sides are collinear, then 
the lines joining corresponding points are concurrent. 

It is noticed that these theorems are simply a specialization 
of the general laws of central projection or perspective. They 
include evidently the laws of affinity as special cases (infinite 
point of concurrence, plane intersections of triangular prisms, 
orthographic and generally parallel projections). 

2. Theorem. — The centers of homology of three homologous 
triangles with the same axis of homology are collinear. 

Let A X B X C X , A 2 B 2 C 2 , A 3 B 3 C 3 be the three triangles whose 
corresponding sides meet in the collinear points a, /?, y. Con- 
sider the two triangles, Fig. 31, A 1 A 2 A 3 and B X B 2 B 3 , then it is 




Fig. 31. 



seen that the lines A X B X , A 2 B 2 , A 3 B 3 are concurrent at 7-. Hence 
the intersections of their corresponding sides are collinear; but 
these points, V ly V 2 , V 3 , are the centers of homology of the given 
three triangles, q.e.d. 

Corollary. — The three triangles A X A 2 A 3 , B X B 2 B 3 , C X C 2 C Z 



82 



PROJECTIVE GEOMETRY. 



have the same axis of perspective; and their centers oj homology 
are the points a, /?, y. Hence the centers of homology of these 
triangles lie on the axis of homology of the triangles A ^^C^ A 2 B 2 C 2 , 
A 3 B 3 C 3 , and conversely. 

3. Theorem. — The three axes of homology of three homologous 
triangles with the same center of homology are concurrent. 

Let A&Cu A 2 B 2 C 2 , A 3 B 3 C 3 be the three triangles with the 
common center of homology V, Fig. 32. Consider the two 




Fig. 32. 

triangles formed by the two systems of lines AJ3 ly A 2 B 2 , A 3 B 3 
and AjCu A 2 C 2 , A 3 C 3 . These two triangles C 3 C^C 2 and 
C"C"C" are in perspective, the line VA 1 A 2 A 3 being their axis 
of homology. Hence the lines joining their corresponding vertices 
are concurrent at S, which proves the theorem. 1 

1 See Casey's Sequel to Euclid, ed. 1900, pp. 77-88. Casey's proofs are based 
exclusively upon metrical properties. 



COLLIN E ATI ON. 83 

4. Theorem. — // in two complete quadrilaterals five pairs oj 
corresponding sides intersect in five collinear points, the point oj 
intersection oj the sixth pair will be collinear with these. 

Suppose that AB, A'B'\ BC, B'C'\ CD, CD'-, DA, D'A'\ 
BD, B'D' are the pairs of sides intersecting on a fixed line s, Fig. 




\ 




;A' 


|8f 

V 

\ 


A i / 

\ \ 1 / 
\ \ I / 

\\ 1 / 

\>/ 

Fig. 33. 





33. Now ABC and A'B'C are homologous triangles, consequently 
AA', BB', CC pass through a fixed point. S. Also BCD and 
B'C'D' are homologous, and as BB' and CC pass through S y 
DD' also passes through S. Hence, as AA', CC, DD' pass 
through S, the triangles ACD and A' CD' are also homologous. 
Now AD and A'D' , CD and CD' intersect in points of s. 
Consequently also AC and A'C intersect on s. This, however, 
is the sixth pair, q.e.d. The line of collinearity may, of course, 
be infinitely distant. 

Ex. 1. Prove dualistically : If, in two complete quadrangles, 
lines joining five pairs of corresponding points are concurrent, 
the line joining the sixth pair is concurrent with these. 

Ex. 2. Consider any three spheres in space which do not 
intersect and exclude each other. Let c ti c 2 , c 3 be their centers 



8 4 



PROJECTIVE GEOMETRY. 



and construct their external common tangent-cones with the 
vertices E v E 2 , E 3 . A common tangent-plane to the cones (E) ly 
(E 2 ) is necessarily also tangent to (E 3 ); hence (E x ), (E 2 ), (E 3 ) 
have the same two external common tangent-planes and E ly 
E 2 , E 3 are therefore necessarily collinear with the line of inter- 
section of these two planes. Similarly it is seen that the internal 
common tangent-cones of (£ x ) and (E 2 ) and of (E 2 ) and (E 3 ) 
have two common tangent-planes which are also common to 
the external tangent-cone of (E ± ) and (£ 3 ). Hence, designating 
the vertices of the internal common tangent- cones by I l9 I 2 , I 3> 
we have the following triads of collinear points: 

E ± E 2 E 3 , EJE 2 I 3 , E 2 E 3 I ly E 3 EJ 2 ; 

i.e , the points of similitude 0} three spheres in space form a com- 
plete quadrilateral. 




Fig. 34. 



In any orthographic projection the spheres are projected into 
circles and the E's and /'s into their centers of similitude; and 
since any three circles in a plane may always be considered as 
the projections of three spheres in space excluding one another, 



COLLINEATION. 



85 



the foregoing proposition also holds for three circles in a plane, 

Fig- 34. 

Ex.. 3. If we now take four spheres in space with the centers 
C 15 C 2 , C 3 , C 4 , and designating the external and internal centers 
of similitude respectively by E ik , /#. for the spheres with the 
centers C\- and Ck, then we find that all external centers of similitude 
lie in a plane. To prove this, remark that there are six centers 
of this kind, E 12 , E 23 , £ 34 , E 41 , E 131 E 24 , and that these are arranged 
in groups of three on six straight lines; they form, therefore, a 
complete quadrilateral and are coplanar, Fig. 35. The centers 




Fig. 35. 

of similitude of three spheres are always coplanar; and since four 
groups of three out of four spheres may be formed, four more 
quadrilaterals of points of similitude may be formed. The 
twelve points of similitude are distributed three by three on six- 
teen axes of similitude. Of the latter, four pass through every 
point of similitude. 



86 PROJECTIVE GEOMETRY. 

If pqrs designates each arrangement of the numbers 1234, 
(pq) the external, (j>q) the internal point of similitude of Cp 
and Cq; moreover, if (pqr) is the axis of similitude passing through 

(M)> (P r )> (<l r )> nnall y (P^ the one throu g n (Pq), (pr), {qr) y 
then the axes may be represented by the following table: 



(234) 


(i34) 


(124) 


(123) 


(134) 


(234) 


(123) 


(124) 


(124) 


( I2 3) 


(234) 


(134) 


( I2 3) 


(124) 


(134) 


(234) 



In this table two axes which belong neither to one and the 
same line, nor to the same column, have always a point of simili- 
tude in common, while this is not true of two axes belonging 
to the same column or line. 

This configuration, which was discovered by Poncelet, is now 
known as Reye's configuration. 1 

§ 27. A Few Applications to Perspective. 

1. Perspective of a Square. — In § 20 it was seen that 
there is always a collineation transforming a quadrilateral into 
any other quadrilateral. The proof of this proposition was 
analytical. In view of its practical application the special case 
is of interest: 

Every quadrilateral may be considered as the central projec- 
tion or perspective of a rectangle or oj a square. 

Let A'B'C'D' be any quadrilateral and L'M'N' its diagonal 
points, Fig. 36. If this is the central projection of a rectangle, 
the line joining M' with N f must be the vanishing line q' ', and 
M f and N r are the vanishing points of the two pairs of parallel 
sides. From this it is clear that the center of perspective joined 
with M' and N f gives two perpendicular lines. In other words, 
the center is situated on a circle having M'N f as a diameter 

1 See Archiv Jihr Mathematik und Physik, 3d series, Vol. I, pp. 124-132. 



COLLIN EA TION. 



87 



(see § 15). Any point on this circle as a center and any line 
i| q f as an axis determine a perspective in which the original 
quadrilateral is the perspective of a rectangle. This rectangle 
can be constructed without difficulty. The quadrilateral may 
also be considered as the perspective of a square. The center O 
must now also be situated on a circle over E'F' as a diameter. 
O is therefore the point, or one of the points, of intersection of 





Fig. 36. 



the circles over M'M' and E'F' as diameters. In § 8 the har- 
monic properties of the complete quadrilateral were obtained 
analytically. Constructing the diagonals of a square, of which 
one is at an infinite distance, those properties appear immedi- 
ately from the square, and as a projection, does not change a 
cross-ratio, it is evident that the same harmonic properties hold 
for the complete quadrilateral. 



88 



PROJECTIVE GEOMETRY. 



2. Perspective of Circles. — Every perspective of a circle 
is called a conic section or simply conic. If two lines are tangent 
to each other at a point A, then the perspectives of these lines 
are tangent at the perspective A f of A. Hence if a circle is 
inscribed to a square, the perspective will give a conic inscribed 
to a quadrilateral. 

The problem oj drawing perspectives oj circles may therefore 
be reduced to the problem of inscribing conies to quadrilaterals. 

By this method the problem can be solved in a simpler man- 
ner than by the ordinary construction from the given circle and 
Qi23B32iP the data of perspective. In Fig. 37 a 
square PQRS and the inscribed circle 
with the points of tangency AA 1 and 
BB 1 have been assumed. Divide OB 
»A and BQ and BP into the same num- 
bers of equal parts and number them 
from O to B and from P and Q towards 
B, starting every time with o. Con- 
nect A with any of the division-points 
IG ' 37 ' on BP, and A x with the corresponding 

point on OB. The point of intersection K of these two rays is 

Q 





Fig. 38. 
a point of the circle. In a similar manner the points of the 
circle in the remaining quadrants may be located. Now the 



COLLIN EATION. 89 

rays from A and A x form two projective pencils, and their prod- 
uct is therefore a curve of the second order. As Z.A X KA is a 
right angle, this curve is a circle (indeed AA 1 Oi= A A Pi, hence 
A^-LAi) and is therefore identical with the assumed circle. 

Now, in order to inscribe a conic into the quadrilateral PQRS, 
touching at AA X BB X , Fig. 38, construct the point O as the inter- 
section of the diagonals PR and QS. Joining O to M and N 
and producing gives AA 1 BB 1 . Applying the same principle of 
bisection by diagonals in analogy with Fig. 37, the proper division 
on OB and PQ is obtained. Having these, the inscribed conic 
is found in exactly the same manner as the inscribed circle. 
The proof of this construction is evident, since every quadrilateral 
may be considered as the perspective of a square, and the per- 
spective does not destroy the projectivity of pencils. 

This construction is very effective in perspective drawing, 
being applicable to all kinds of quadrilaterals. 

// is also remarked that conies (perspectives 0) circles) are 
curves 0] the second order. 

This idea will be fully discussed in the next chapter. 



90 



PROJECTIVE GEOMETRY. 



§ 28. Exercises and Problems. 

1. Given two straight lines and a point; to draw a straight 
line through this point passing through the inaccessible point of 
intersection of the given lines. 

Solution. — Let g and h be the given lines and A the given 
point. Through A draw any two lines cutting g and h in PQ 
and RS, Fig. 39a and 396. Join PS and QR and produce till 
they intersect at M. .Designating the inaccessible point by N, 




Fig. 39a. 



PQRS may be considered as the perspective of a square having 
M, N, A as diagonal points. Hence any third line through M 
cutting g and h at T and V is the perspective of a line parallel to 
QR and PS. From this it follows that the line joining A to the 
point of intersection B of TR and QV is the required line. 

2. Inscribe a conic within a rectangle; within a trapezium; a 
rhombus. 

3. Draw two homologous quadrilaterals. 



COLLI NEATION. 



91 




B^r- 



Fig. 396. 



4. Draw the perspective of a cube; of a cylinder; of a pyramid; 
of a hexagonal prism. 

5. Draw the perspective of two concentric circles. 



CHAPTER III. 

THEORY OF CONICS. 

§ 29. Introduction. 

The Greeks originally studied conies as plane sections of 
cones. 1 Steiner and Chasles considered them as products of 
projective pencils and ranges, denned by anharmonic ratios, 
von Staudt and Reye, however, define this relation purely by 
harmonic division. I shall follow Steiner' s method, by which 
the projective properties of the circle (see § 12) are easily ob- 
tained and transformed to conies by central projection. Follow- 
ing this method, it becomes necessary to show that all curves of 
the second degree as obtained by projective pencils and ranges 
are also produced by plane sections of cones, or as perspective 
collineations of the circle. 

Conversely, it must be shown that every curve of the second 
degree may be projected into a circle. This is the method followed 
by Poncelet, Steiner, and a majority of modern writers on pro- 
jective geometry. From a purely geometrical standpoint von 
Staudt's and Reye's methods are to be preferred. 



1 Menachmus obtained conies as intersections of planes perpendicular to the 
elements of a right cone. In case of an "acute-angled cone" (opening angle at the 
vertex < oo°) the conic was called ellipse; of a "right-angled cone" (angle at 
vertex = Qo-°) a parabola; of an "'obtuse -angled cone*" (angle at vertex > oo°) an 
hyperbola. Apollonius, who introduced these names, extended the proofs to 
oblique cones. 

92 



THEORY OF CONICS. 93 



§ 30. Identity of Curves of the Second Order and Class and 

Conies. 

The general equation of a circle is 

(1) (x-a) 2 +(y-b) 2 = r 2 . 

To obtain the equation of this circle in line- coordinates, 
assume the equation of its tangent in the form 

(2) ux-\-vy-\-i=o. 

This represents a tangent if the distance of the center (a, b) 
from the line (2) is r; i.e., if 

au bv 1 



\Zu t +v 2 \/u 2 +v 2 \/u 2 -\-v 2 
or 

(3) r 2 (u 2 +v 2 )-(au+bv+ i) 2 = o. 

Every pair of values u, v satisfying (3) gives the line-coor- 
dinates of a straight line tangent to the circle (1). Equation (3) 
represents, therefore, (1) in line-coordinates (see §6). Both (1) 
and (3) depend upon three essential parameters. The formulas 
for perspective in their dual interpretation each depend upon 
three essential parameters, rlence, applying a perspective to 
either (1) or (3), i.e., to the given circle, we can in both cases 
dispose of the six parameters in such a manner that the trans- 
formed equations assume any given form of the second degree 
in x and y, or in u and v. This means that every curve of the 
second order or class may be considered as a conic section {per- 
spective 0] a circle). 

Conversely, if the general equation of a conic of the second 
degree is given, it is always possible to find a perspective which 
will transform this equation into that of a circle. Hence every 
curve of the second degree may be projected perspectively into a 
circle. Curves 0) the second degree and conies are therefore identical. 



94 PROJECTIVE GEOMETRY. 

Ex. i. The perspective transformation 



j x 



lx-\-my J \-n > 



y- - ^ 



lx-\-my-\-n 

transforms the general equation 

ax' 2 -\- by' 2 + cx'y' + 2dx' + 2ey' + / = o 
into 

x 2 (a-{- 2dl+ jl 2 ) + y 2 (h+ 2em-\- )m 2 ) + xy(c-\- 2dm-\- 2el+ 2Jlm) -f 

x(2dn+ 2 fin) + v(2^w+ 2/ww) + ]n 2 . 

Find the values of / and m which will transform the given 

equation into that of a circle. 

Ex. 2. Solve the dual problem of Ex. i. 

Ex. 3. Find a circle and a perspective, so that the perspec- 

x 2 y 2 
tive of the circle is the ellipse -5+ -r- 2 = 1. 

Ex. 4. Given the circle x 2 -\-y 2 =i. Transform this circle 
by the perspective 

x' y 

x ~y-i> y ~y-i 

Discuss the result geometrically and show that the center of 
the circle is the focus of the transformed circle. 



§31. Linear Transformation of a Curve of the Second Order. 

1. By the translation 



x = x / +a ly 

>y=y'+h 



. .. THEORY OF CONICS. . 95 

the equation 
(2) ax 2 -\-2bxy-\-cy 2 +2dx+2ey-{-j=o '/ 

is transformed into 

ax' 2 + 2bx f y+c'y 2 +2(aa 1 -{-bb 1 +d)x f + 2(ba t skcb l ± e)y'+ 

aa 1 2 + 2ba i b 1 + cb l 2 + 2da 1 + 2eb 1 -\- /=o. 

In order that the coefficients of x f and y' disappear, a x and b x 
must be chosen, so that 

aa l +bb 1 J rd = o, 
ba^cb^e^o. 



If ac— b 2 ^o, we find for a t and & t the values 

(3) 



cd bd—ae 

0l = 



ac-b 2 ' 

By this assumption the transformed equation reduces to (the xf 
and / being replaced by x and y) 



(4) 

where 



ax 2 -\-2bxy-\-cy' 



ac—b 2 



o ; 



J = d(be- cd) + e(bd- ae) + /(ac- b 2 ) = 



a b d 
bee 
d e j 



2. Turning the coordinate axes now through an angle 6; i.e., 
making the transformation 



(5) 



j x=x/ cos 6—y sin (9, 
I ^ = ^ sin d+y cos #, 



and again suppressing the primes of x and y } the result is 



Ax 2 +2Bxy+Cy 2 +- 



dc — b' 



= 0, 



96 PROJECTIVE GEOMETRY. 

where 

A — a cos 2 0+ 26 sin 6 cos 0+ c sin 2 d, 
2 B= (c—a) sin 2^+26 cos 20, 
C = a sin 2 # - 2& sin cos d+ c cos 2 0. 

Choosing d so that B = o } or 

2& 

(6) tan 2 # = — -, 

the transformed equation reduces to 

A 



(7) Ax 2 +Cy 2 = 



b 2 —ac 



To determine A and C, we have from the foregoing expressions 
for 4, 2B, C: 

K ' (B 2 -A-C = b 2 -ac. 

But when is chosen so that B = o, A -C = ac—b 2 . Hence A and 
C are roots of the equation 

z 2 — (a+c)z+ac—b 2 = o. 

If now b 2 —ac^o, two cases, b 2 —ac>o and b 2 —ac<o> must be 
distinguished. 

It is further assumed that J^o. In the first case, b 2 —ac>o, 

it follows, since now AC = ac— b 2 < o, that A and C are of different 

J 
sign. Hence, no matter what the sign of J, . , 2 _ — - and 

-^-tt^ : have different signs and are always real. The equa- 
tion therefore represents an hyperbola. If, in addition to b 2 — ac >o, 
J = o, then the equation may be resolved into two linear factors; 
the hyperbola degenerates into two intersecting straight lines. 
In the second case, b 2 —ac < o,A -C = ac- b 2 >o. Both A and 



THEORY OF CONICS. 97 

C have the same sign; hence also j-t^ r and zttt- 2 r have 

/x \0 CLC ) (_• ( — Q/C) 

the same sign. According as this is positive or negative, the 
equation represents a real or an imaginary ellipse. If J = o, the 
ellipse degenerates into two intersecting imaginary lines. 

3. Finally the case b 2 —ac = o must be considered. Here 
b = ±\Aic. Considering the case b = +Vac, the general equa- 
tion (2) reduces to (coordinates x f ) y f ) 

(9) (Vaxf -f V O'') 2 + 2dx? +2ey f + j = o. 

Putting \/a = r sin 6, Vc = r cos 6; i.e., r 2 = a+c, tan = -v|— i 
and dividing the whole equation by r 2 , we have 

(10) (x? sind + y cosd) 2 + -^(2dx'+2ey' + })=o. 

Turning the coordinate axes through an angle Q\ i.e., putting 

x = x t cos Q—y' sin 6, 
y=x! sin0+/ cos 0, 

or x*=x cos d-\-y sin d, 

y=? — xsmd-{-y cos d, 

V a \fc 

equation (10) becomes, after putting sin = , ., cos 0= , - > 

Va+c Va+c 

and reducing, 

(11) y 2 + * —^ \(dVc-eVa)x+ 

(a+c)Va+c( 

(dVa+eV^y+hjVa+c [ =0. 

Making finally the translation, by replacing y and x by y+ft 
and x+a respectively, we have 



9 8 



PROJECTIVE GEOMETRY. 



(12) y 2 +— — 77= ](dVc-eVa)x+. 

(dVa+eVc-{-j3(a+c)Va-\-e)y+a(d\/c—e\/a) + 



—(a+c)Va+c+ijVa+c 



Letting 



(12) becomes 



n eVc+dVa 

p = and 

(a+c)Va+c 



a = — 



(eVc+d\/q) 2 +j(a+c) 2 
2 (dVc— eV a) (a + c)Va+ c * 

dVc—eVa 



»2_ 



■*i 



which represents a parabola. When b 2 —ac = o, then 
a & d 



J = 



bee 

d e f 



(cd 2 —2bde+ae 2 ). 



But b= +Vac, hence 

J = — (ae 2 — 2V ac • de+ cd 2 ) = — (eVa— dVc) 2 ; 



hence 

(13) 



:X. 



{a-\- c)\/ a+'c 

If in (11) dVe—e\/a = o y i.e., J=o, then the equation may be 
resolved into two linear factors in y only, and represents con- 
sequently two parallel lines. These are real and distinct, coinci- 
dent (real), imaginary and distinct, according as 

(dVa+eV7) 2 - j(a+e) 2 ^o. 

The case b = —Vac may be treated in a similar manner and 
leads to no new result. 

4. From the foregoing short discussion it is seen that the 
character of the general equation of the second order may be 
established by means of translations and rotations; i.e., by special 



THEORY OF CONICS. 



99 



cases of the linear transformation. In this, two algebraic expres- 
sions between the coefficients are of fundamental importance, 

be. 

, which for abbreviation we may designate 



namely, b 2 — ac = 
by t, and 



a b 



A = d(be- cd) + e(bd- ae) + j(ac- b 2 ) = 



> 



a b d 
bee 
d e } 



According as : = o the general equation represents an hyper- 
bola, a parabola, or an ellipse, if A ^o. When A = o these 
curves degenerate into intersecting, parallel or coincident, or 
imaginary intersecting lines. The determinant A is called the 

discriminant of the equation. We may call z = b 2 —ac = , 

the characteristic determinant, or simply characteristic. 

We shall now show that the discriminant and characteristic 
of an equation of the second order are not changed by a translation. 
By the translation in which x and y are replaced by x+a x and 
y+b ly equation (2) is transformed into 
(14) ax 2 -\- 2bxy+ cy 2 + 2(aa 1 -\- bb t + d)x-\- 2{ba x -\- cb 1 -\-e)y 

+ a 1 (aa 1 + bb^ d) + & 1 (5a 1 + cb x -{- e) + da x -\- eb t +}=o. 
From this it is apparent that z has remained invariant. The 
discriminant 

a b aa t +bb ± +d 

b c ba^cb^-r-e 

■bb x +d ba^+cb^+e a l (aa 1 +bb 1 +d) + b l (ba 1 +cb 1 +e) 

+ da 1 + eb l +j 

a b d 

bee 

aa x +bb x +d ba x + cb x +e da^ eb x + f 

a b d 

bee 

d e / 

which is the original discriminant. 



A = 



L.ofC. 



IOO PROJECTIVE GEOMETRY. 

In a translation z and A are invariants. The same is true 
of z and A in a rotation, and consequently in any motion of the 
plane in itself. No special proof for the case of rotation will 
be given, since it is contained in the linear transformation which 
will now be treated. 

5. A linear transformation (§19, XII) 

(I5) \y= rx ' + dy' + r j , 

leaves the line at infinity invariant. It may therefore be ex- 
pected that such a transformation does not materially change 
the expressions z and A. 

As the constants $ and t) mean simply a translation in addi- 
tion to the special linear transformation 

\x=ax'+py', 

and as a translation does not change z and A, it is sufficient to 
study the effect of (16) upon the general equation (2). Making 
in (2) for x and y the substitution (16) and afterwards replacing 
xf and y by x and v, we have 

(17) (aa 2 +2bar+cf)x 2 +2\aaP+b(ad-\-j3r) + crd}xy 

+ (ap 2 +2b(3d+cd 2 )y 2 +2(da+e r )x+2(d(3+ed)y+} = o. 

The discriminant of this equation is 

aa 2 +2ba]-+cf aa@+b(ad+Pf) + crd da+ey 
aa(3+b(ad+p r ) + c r d ap 2 + 2 bpd+cd 2 dp+ed 

da+ey d{3+ed } 



which reduces down to 



(18) J' = (ad-p r y 



a b d 
bee 
d e J 



(ad-p r ) 2 4. 



THEORY OF CONICS. 10 1 

In a similar manner there is found 

(l 9 ) T' = (6^-/? r ) 2 T. 

From this it follows that the character of a curve 0} the second 
order is not changed by a linear transformation. 

It is always assumed that ad—fty^o. Indeed, according 

asr = o, also t' = o; and also as i = o, J'=o. 

For a rotation, t' = (cos 2 #+sin 2 0)t = t, 

i.e., in this case r and A are invariants. In the next section it 
will be seen that conies are characterized by their pole- and 
polar involutions on the line at infinity (involution of diam- 
eters). 

Thus, it is also geometrically evident that in a linear trans- 
formation, which does not change the involution of the infinitely 
distant line, the type of a conic is not changed. 

Ex. 1. Find (18) from the preceding unsolved determinant. 
Calculate also x r . 

Ex. 2. If b 2 —ac = o, assume b^—Vac and transform, with 
this condition, the general^ equation (2) to the normal form 
y 2 = 2px. 

Ex. 3. Discuss the curve determined by y 2 — 2xy-\-x 2 — 1 =0. 

Ex. 4. What curve does the equation 

ax 2 + (a+ b)xy+by 2 + (a+ c)x+ (b + c)y + c = o 
represent ? 

Ex. 5. If in the linear transformation ad—ftr=o i i.e., tt = -*, 
we have 



-*(£*+•)• 



102 PROJECTIVE GEOMETRY. 



X B 

Hence, no matter what the values of x? and V may be, - = % = 

J - y 

constant. The whole ^/-plane is transformed into the straight 
line xd—yft = o. 



§ 32. Polar Involution of Conies. Center. Diameters. Axes. 

Asymptotes. 

1. In §§ 12 and 13 the involutoric properties of the circle 
have been explained. As a collineation does not change pro- 
jective properties, it is clear that the following theorems hold 
for conies. (As the figures of involution referred to in this sec- 
tion are in close analogy with those of the circle, their reproduc- 
tion is left to the student.) 

I. The polar s of the points of a straight line I pass through 
a fixed point L, the pole of I. 

II. The poles of the rays of a pencil P lie on a straight line p, 
the polar of P. 

From this follows immediately 

III. // the pole L of a straight line I lies on a second line g, 
then the pole G of g lies on I. 

IV. // the polar I of a point L passes through the pole G of a 
second line g, then the line g passes through the pole of I. 

Consider now any straight line / and its pole L. Let a be 
any ray through L cutting / at A. The pole A 1 of a lies on /. 
Hence the pole of the ray a x , passing through L and A„ coin- 
cides with A. Taking any number of rays a, b, c, d, . . . and 
constructing as before their corresponding rays a lf b x , c u d ly . . . , 
a system of coincident polars and poles through L and on / are 
obtained which 'are in involution; i.e., every pair of correspond- 
ing rays and poles, xx ± and XX t , are harmonic with the double- 
rays and double-points through L and on /, respectively. The 
double-elements are real when / intersects the conic really; i.e., 
when L admits of two real tangents to the conic. If / does not 
intersect in real points, then the involution has imaginary double- 
elements. In case that / is tangent to the conic, the double- 



THEORY OF CONICS. 103 

elements are coincident. The points of intersection of / and 
the tangents from L to the conic, whether real, coincident, or 
imaginary, give in all cases the double-elements of the involu- 
tion. .Accordingly, hyperbolic, parabolic, and elliptic involu- 
tions are distinguished. 

Two corresponding rays x, x lt and I, with their poles X, X v and 
L always form a self- polar triangle; i.e., a triangle zvhose vertices 
are the poles of its opposite sides. 

All these properties might be derived directly from the gener- 
ation of conies by projective pencils and ranges; i.e., without 
reference to the perspective of the circle. We have used this 
method to lay particular stress upon the invariance of these 
properties by projective transformations. Examples will be given 
later on to show how some of the propositions (all, for that matter) 
in this connection may be derived independently of perspective. 

2. It is now of the greatest interest to investigate the involu- 
tions of poles and polars when the latter are assumed in special 
positions. Let / be at an infinite distance. Then for every ray 
a through L cutting the conic at P and P 1 (LAPP 1 ) = — i = 
(L 00 PP X ), or LP= —LP V Every ray through Z, the pole of the 
line at infinity, therefore cuts the conic in two points which are 
equally distant from L. This point is therefore called the center 
of the conic. 

To every ray through the center corresponds an infinitely dis- 
tant pole. Call the center O. Of great importance is the polar 
involution through O. The poles A and A 1 of any two corre- 
sponding rays through O are infinitely distant. Any ray through 
A cutting a x in B and the conic in C and D is parallel to a, and 
(BACD) = (B*CD) = -i; hence BC=-BD; i.e., B bisects CD. 
Two corresponding rays of the involution through O are called 

CONJUGATE DIAMETERS of the Conic. 

Including imaginary elements, the foregoing properties of the 
polar involution at the center give the following theorems : 

V. All chords of a conic parallel to a diameter are bisected by its 
conjugate diameter. The relation between conjugate diameters is 
reversible. 



104 PROJECTIVE GEOMETRY. 

VI. // a diameter intersects a conic, then the tangents at the 
points of intersection are parallel to the conjugate diameter. 

3. Definitions. — The rectangular pair of the polar involution 
at the center are called the axes of the conic. 

The double-rays of the involution of diameters of a conic are 
called the asymptotes. 

The points at which the polar involution is rectangular are 
called the foci of the conic. 

In these definitions it is assumed that the involutions exist. 
From the definition of a self-polar triangle it is easily concluded 
that the involutions on two of its sides are hyperbolic, while on the 
third it is elliptic. If we now consider a self-polar triangle hav- 
ing the center O of the conic as one of its vertices, then two dis- 
tinct cases may occur. 

First. The involutions of poles on two conjugate diameters may 
both be hyperbolic, while the polar involution at the center, con- 
sequently the involution of poles on the line at infinity, is elliptic. 
The involution of conjugate diameters has no real double-rays; the 
conic is an ellipse. 

Second. The involution at the center is hyperbolic; it is hyper- 
bolic on one diameter and elliptic on the conjugate diameter. The 
involution of conjugate diameters has real double-rays; i.e., the 
conic has real asymptotes and is an hyperbola. 

From theorem V it follows immediately that the ellipse and 
hyperbola are symmetrical with respect to both axes. 

The existence of the ellipse, hyperbola, and a special involu- 
tion of diameters in connection with a conic, called parabola, will 
be proved in the next section. At the same time the existence 
of foci will be proved. 

§ 33. Existence of Ellipse, Hyperbola, Parabola, and their 

Foci. 

1. In Fig. 40, just as in Fig. 19, § 15, let 5 be the axis, <f and 
r the counter-axes, and C the center of collineation. Assume any 
circle K with C as a center and determine the pole O of r with 



THEORY OF CONICS. 



I05 



respect to K. Draw also the polar involution at O, of which OM 
and OM ± are a corresponding pair intersecting K at X, Y and 
X 1} Y,, respectively. Now (OMXY) = (OM 1 X 1 Y 1 ) = -i. In 
the central projection r and consequently M and M x are projected 




Fig. 40. 



to infinity. Hence, designating the projected figure by primes, 
(0 , ooX , F , ) = (0 , ooX/F/) = -i; i.e., O'X'^-O'Y', 0'X/ = - 
0'YI\ O' is the center of the transformed circle, and the polar 
involution at O' becomes the involution of diameters. Now the 
polar involution at O is elliptic, parabolic, or hyperbolic accord- 
ing as r does not cut K, touches K, or intersects it in two points. 
The same is evidently true of the involution of diameters at O r . 
In case that r is tangent to K, O coincides with the point of tan- 
gency of r; O' is projected to infinity, so that the diameters be- 
come all parallel. This is the case of the parabola. A parabola 
may therefore be considered as a conic tangent to the line at 
infinity. With the existence of these different involutions of 
diameters the existence of the ellipse, the parabola, and the hyper- 
bola is proved. To sum up, the central projection 0} a circle is an 



106 PROJECTIVE GEOMETRY. 

ellipse, a parabola, or an hyperbola according as it does not intersect 
r, is tangent to r, intersects r. Further, a conic is an ellipse, 
a circle, a parabola, or an hyperbola according as the involution oj 
diameters is elliptic, elliptic and rectangular, parabolic (parallel 
diameters with infinite center), or hyperbolic. 

2. Consider now the rectangular polar involution at the center 
O of the circle K. The central projection of O coincides with 
itself, and for the corresponding pair of two perpendicular diam- 
eters AB and BE, A'B'±B'E'. The points of intersection P and 
Q of DE and AB with a' are the poles of A'B' and D f E r . The 
vanishing line q f is therefore the polar of C with regard to K f . 
The same holds for any pair of perpendicular diameters of K and 
their transformations. 

The involution oj polars at C oj K r is therefore rectangular; 
C is a focus oj K! . 

Any conic which is the perspective oj a circle with the center 
oj perspective as a center has this center as a focus. 

The construction also shows that a jocus lies on the major axis 
oj the conic. 

Ellipse and' hyperbola are symmetrical with respect to their 
axes; both curves have therefore two foci (real). That a conic 
cannot have more than two real foci is seen from the construc- 
tion of Fig. 40, and also from the fact that every point not on 
the axes admits of oblique pairs of polars. 

The double- rays of the rectangular polar involutions at 
the foci pass through the circular points at infinity; they may 
be considered as imaginary tangents to the conic from its 
foci. The foci of a conic may therefore also be defined as 
follows : 

The foci oj a conic are the points of intersection of the tangents 
from the circular points at infinity to the conic. 

Two of these intersection-points are real, the other two are 
imaginary and, on account of the symmetry, are necessarily situ- 
ated on the other axis. Ellipse and hyperbola admit, therefore, 
also of two imaginary foci. 

Analytically this also appears by writing the equations of 



THEORY OF CONICS. 107 

the imaginary double-rays at the foci, when their distance from 
the center of the conic is c: 





1. 


(x—c) + iy = o. 




2. 


(x—c) — iy = o. 




3- 


(x+c) + iy = o. 




4- 


(x+c) — iy = o. 


From 1 and 2, 




x=c\ y=o. 


From 3 and 4, 




x= — c; y=o. 


From 1 and 4, 




x=o; y= —ic. 


From 2 and 3, 




x = o; y=+ic. 



The solutions of 1 and 3, 2 and 4 give the circular points. 

§ 34. Construction of Foci Independent of Central Projection. 

To the pencil T of parallel rays a, b, c, . . . considered as 
polars of the conic K, Fig. 41, correspond the poles A', B', C, . . . 
on the conjugate diameter n of the direction of T. To the diam- 
eter m || T corresponds as pole the infinitely distant point of n. 
To the line at infinity belonging to the pencil T corresponds 
as pole the center /' of K. 

From A\ B', C, . . . , s i 7 , draw rays a', &', c', . . . , f per- 
pendicular to m. These rays form another pencil 5 of parallel 
rays which is projective to the pencil T. As these two pencils 
are perpendicular, their intersection A u B u C 19 . . . , M x is an 
equilateral hyperbola, having m and its perpendicular through 
/' as asymptotes. Both pencils T and S intersect the axes each 
in two coincident projective ranges, for instance 

(ABC ...)a(A 2 B 2 C 2 ...). 

On this axis, to the point /' of the first range corresponds 
the infinitely distant one on the same axis. If V is taken as a 
point of the second range, then its corresponding one is infinitely 
distant. Hence /' and the point at infinity on the horizontal 



io8 



PROJECTIVE GEOMETRY. 



axis may be interchanged without disturbing the projectivity; 
the foregoing point-ranges form therefore an involution (§ 3). 
The double-points of this involution are the points where the 
equilateral hyperbola H cuts the axis. To the ray / (pencil T) 
through one of these points, say F, corresponds the perpendicular 
ray /' in the pencil S. But every ray of 5 passes through the 




M" 



Fig. 41. 



pole of the corresponding ray in T. Hence /' contains the pole 
of /, and / and /' form consequently a rectangular pair of the 
polar involution around F. The polar involution of any point 
on an axis contains, however, another rectangular pair, namely, 
the axis itself and the perpendicular to it through the given point. 
The polar involution about F has therefore two rectangular 
pairs and is consequently itself rectangular (§ 5). The point 
F, according to definition, is therefore a focus. As there are 
two double-points of the involution (ABC . . .) 7\(A 2 B 2 C 2 . . .), 



THEORY OF CONICS. 109 

there are also two foci. Assuming that the points of a pair AA 2 
are on the same side of the center of involution, then 

I'A-FA t = + k\ 

and FF = k ; the foci are real. But on the other axis FA * • FA 2 * = 
— k 2 , i.e., the double-points of the involution, or the foci, are imag- 
inary. This is in accordance with the statement in the foregoing 
section. 1 

Ex. 1. Carry out construction of this section on a large sheet. 

Ex. 2. Instead of taking an ellipse for the conic K, take an 
hyperbola. 

§ 35. Focal Properties of Conies. 

1. From Fig. 40, § 33, we have, if R designates the point of 
intersection of PD with r (not shown in the figure), 

(CoDDR) = (CPD f o C ) l 
or CD:CR = CD' :PD'. 

Designating the distances of C and D' from r and q f by y and d, 
respectively, there is 

QR:PD' = r :d=CD:CD'\ 

consequently 

CD' CD 

~^~ = = constant. 

This result may be stated by the theorem: 

The ratio between the distance oj any point of a conic from one 
of its foci and the distance of the point from the polar of this focus is 
constant. 



1 Since an involution of right angles does not admit of real double-rays, it 
follows that the foci are within the conic; i.e., within that portion of the plane 
from which no real tangents may be drawn. They are situated on an axis, since 
in any other case the polar involution would have oblique pairs. 



no 



PROJECTIVE GEOMETRY. 



Definition. — The polar of a focus o) a conic is called directrix. 

This theorem, as well as its converse, may be used to define 
conies, as was done by Pappus (Mathematical Collections) : 

The locus oj a point whose distances from a fixed point and a 
fixed straight line (not passing through the fixed point) have a con- 
stant ratio is a conic. The fixed point is the focus, the fixed line 
the corresponding directrix. 

CD 

If, in Fig. 40, K does not intersect r, i.e., if < 1, K' is an 

CD 

ellipse ; if K touches r, = 1 , and K f is a parabola ; if K inter- 

CD 

sects r, >i, and K r is an hyperbola. The figure may easily 

be drawn for the case of an hyperbola or a parabola. 




Fig. 42. 



2. From Pappus' metrical definition a number of properties 
of conies may be derived. Given a conic K and its foci F and F u 
Figs. 42 and 43, according as we take for K an ellipse or an hy- 



THEORY OF CONICS. 



in 



perbola. Both curves are symmetrical with respect to both axes. 
/ CD " 



"he ratios I ) are therefore the same for both foci and their 

corresponding polars (directrices). Taking any point A on K and 
designating the focal distances AF and AF X by r and r v and the 
distances, from the corresponding directrices by d and d lt we have 



r r x 

~3 = ~T = constant (Pappus). 



From this 



r±r 1 
d±d x 



= same constant as above. But in an ellipse 




Fig. 43- 

d+ d x = constant, and in an hyperbola d— d x = constant. Hence 
the theorem: 

The sum of the radii vector es {AF, AF t ) of any point of an 
ellipse is constant. 

The difference of the radii vectores of an hyperbola is constant. 
In both cases the constant is equal to the distances of the vertices of 
the curves. 



112 PROJECTIVE GEOMETRY. 

The second part of this theorem results by taking A in one 
of the vertices. In case of an ellipse we have then 

r+r, r r 

in case of an hyperbola 

r _ r — r — -d 

a 

3. Given a conic K, Fig. 44, and its foci F and F v Take 
any point P in the plane of K and construct the polar involution 
around P and its rectangular pair PR ly PR. Connecting P 
with all pairs of the involution on the axis which are formed by 
couples of rectangular conjugate polars of K parallel to PR 




and PRi (see Fig. 41, § 34), an involution of rays at P is obtained 
in which PF and PF f are the double-rays, and PR, PR^ the 
rectangular pair. PR and PR± are consequently the bisectors 
of the angles formed by PF and PF V In the polar involution 
around P, the tangents PU and PV from P to K are the double- 
rays and, according to the construction, PR, PR^ the rectangular 
pair. The angles formed by PU and PV are therefore also 
bisected by PR and PR V Hence the theorem: 



THEORY OF CONICS. 



IJ 3 



The angles formed by the tangents {real) from a point to a conic 
are bisected by the bisectors of the rays joining this point to the 
foci; or these tangents form equal angles with the focal rays (PF, 

PFi). 

If P lies on K, say at Q, then PU and PV coincide with the 

tangent / at Q. We have therefore the corollary: 

The tangent at any point of a conic includes equal angles with 

the focal radii at this point. 




Fig. 45. 



4. If in Figs. 42 and 43 F X A is produced and AB made equal 
to AF (F 1 B = 2a, the major axis), then BF±AC (tangent at A); 
hence BC = FC. As OC \\F 1 B = iF 1 B, we have the theorems: 

The locus of the reflected images 0} a focus on all tangents 
of an ellipse or an hyperbola is a circle having the other focus as 
a center and the major axis as a radius. 

The locus of the foot-points of all perpendiculars from the foci 



ii4 



PROJECTIVE GEOMETRY. 



of an ellipse, or an hyperbola, to their tangents is a circle over the 
major axis as a diameter. 

In case of a parabola, Fig. 45, the first circle becomes the 
directrix f 1 and the second the tangent v at the vertex. Suppose 
D to be a point where two perpendicular tangents of the parabola 
meet, and let A and A' be the points of tangency. We have 

a + a'=— ; hence AF t and A'F X include an angle of n. D is 

therefore the pole of a focal chord A A' and, as such, lies in the 
directrix. Hence : 

The tangents front any point of the directrix of a parabola to 
the parabola are perpendicular to each other. 

It is known from § 33 that the polar involutions around a 
focus are rectangular. Thus, if A A' is a focal chord and A t its 
pole on the directrix, FA x ±AA f at F. The foregoing statement 
is therefore only a part of the general proposition, since ZAFA l = 
right angle, Fig. 46: 




Fig. 46. 

The portion of a tangent of a conic between its point of tangency 
and its intersection with the corresponding directrix appears as 
subtended by a right angle when seen from the focus. 



THEORY OF CONICS. 115 

Further, from the fact that the polar of a point which is situated 
on another polar passes through the pole of this polar, and that 
the polar involution at a focus is rectangular it follows, Fig. 46: 

The rays pining a focus with the point of intersection of two 
tangents and the point of intersection of the chord of contact of 
these, tangents with the corresponding directrix are perpendicular. 

Ex. 1. If from any foint O at a distance c from the center 
of a circle with radius r two perpendicular secants are drawn, 
intersecting the circle in the points A, B, C, D, then 

OA 2 +OB 2 +OC 2 +OD 2 = 4r 2 . 

Assuming this proposition, prove: 

The locus of the point of intersection of two tangents to an 
ellipse or an hyperbola which cut at right angles is a concentric 
circle. If a and b are the parameters (major and minor half- 
axes), the radius of the circle is \/a 2 ±b 2 . 

Ex. 2. Let Q and R be the points of intersection of a third 
tangent with the two tangents to a parabola from a point P. 
Prove that ZRPF is equal to the angle which QP makes with 
the diameter of the parabola through P. 

Ex. 3. Applying the proposition established in the fore- 
going exercise, prove that the circle circumscribing a triangle 
formed by three tangents to a parabola passes through the focus. 

Ex. 4. The locus of the foci of all parabolas which touch the 
three sides of a given triangle is the circumscribing circle of the 
triangle. (Cremona.) 

Ex. 5. Suppose a quadrilateral A BCD is circumscribed 
about a conic with the points of tangency KLMN, Fig. 47. The 
pairs of sides AB and CD, BC and AD, CA and ED intersect 
each other in the three points OPQ. The pole of AC is the 
point of intersection of KN and LM and is the point of con- 
currence of PO, ML, DB, NK. Similarly, KL, AC, NM meet 
at a point of PO, the pole of BD. Hence, in a quadrilateral 
circumscribed to a conic, the diagonals form a self-polar tri- 
angle. If A, B, N, and L are given, then we may choose Q 






n6 



PROJECTIVE GEOMETRY. 




Fig. 47- 



THEORY OF CONICS. 



117 



at random on NL, by which the fourth side CD and its point 
of contact is perfectly determined. From this the proposition 
follows : 

Let PAB be a triangle circumscribed to a conic, and LN the 
chord 0) contact of the tangents PB and PA; then the lines join- 
ing any point Q on LN io A and B are conjugate polars. 

Ex. 6. Prove: The portion of a movable tangent of a central 
conic between the two tangents at the vertices subtends right angles 
at the foci. 

Ex. 7. The lines joining the points of intersection of all circles 
through the foci with the tangents at the vertices of a central conic 
are the tangents of this conic. (The two points joined must not 
be equally distant from the axis.) 

Ex. 8. Consider again, Fig. 48, two tangents intersecting 
at P, and let their chord of contact AB intersect the directrix 




Fig. 48. 



at C. Then PF is the polar of C. Consequently FAPBC is 
a harmonic pencil in which one pair, FP, FC is perpendicular. 
FP therefore bisects the angle AFB, § 5. Hence the proposi- 
tion: 

The line joining a focus to the point of intersection of two tan- 



Il8 PROJECTIVE GEOMETRY. 

gents of a conic bisects the angles formed by the rays joining the 
focus to the points of contact. 

Ex. 9. Consider two fixed tangents and a movable tangent 
of a conic, and join a focus to their points of intersection and 
points of contact. Applying the proposition of Ex. 8, prove that 
the piece of a movable tangent included by two fixed tangents appears 
under a constant angle from a focus. 

In case of an hyperbola whose asymptotes include an angle <£, 
the above constant angle with reference to the asymptotes as 

fixed tangents is n . 

Ex. 10. The extremities of a tangent, determined by the asymp- 
totes, and the foci of the hyperbola are concyclic. 

By means of this proposition it is easy to construct an hyper- 
bola by its tangents when the asymptotes and the foci are known. 



§ 36. Analytical Expression for Tangent and Polar. 

1. Although problems connected with tangents and polars 
of general conies have so far been simply treated without their 
analytic forms, it is of great value for the developments that will 
follow to establish their equations by means of transversals and 
anharmonic ratios. Let (x v y x ), (x 2 , y 2 ) be two points A and C; 
then the coordinates of any point B of the straight line AC are 
given by the equations 



x t — Xx 2 ?i — ^2 , . AB 

to x =^r> y=^=T> where x =cb- 



Substituting these values in the general equation 

u=ax 2 +2bxy+cy 2 +2dx-{-2ey-\-f=o, 

and multiplying by (1 — A) 2 , we obtain, after arranging according 
to ascending powers of A, 



THEORY OF CONICS. 119 

(2) U 1 —2Xv+X 2 U 2 = o, 

where 

u 1 = ax 1 2 -{- 2bx 1 y l + cy x 2 + 2dx 1 -\- 2ey x -\- /, 
u 2 = ax 2 2 + 2bx 2 y 2 -\- cy 2 2 + 2dx 2 + 2ey 2 +}, 
v = x x (ax 2 + by 2 + d) + y x (bx 2 + cy 2 + e) + dx 2 + ey 2 + ) 
= x 2 (ax x + by x +d) + y 2 (bx x + cy x + e) + dx x +ey x + f. 

From (2) two values of / are obtained w T hich w T hen substituted 
in (1) give the coordinates of the points of intersection of AC 
with the conic U. In case that these two points coincide, the 
roots of (2) will be equal and^4C is a tangent to U. Now, the 
condition for equal roots is 

(3) v 2 -u x u 2 =o. 

Suppose that (x 2 , y 2 ) itself is on U, then u 2 = o, and the con- 
dition reduces to v=o. Every point (x x , y x ) which satisfies v=o 
lies on the tangent at (x 2 , y 2 ). The equation of the tangent at 
(x 2 , y 2 ) is therefore 

(4) x(ax 2 + by 2 + d) + y(bx 2 + cy 2 + e) + dx 2 +ey 2 +f = o. 

2. Let A and B be the points (x x , y x ), (x 2 , y 2 ) and C, D the 
points of intersection of the straight line AB with U, correspond- 
ing to the roots of (2). If A, B and C, D are two harmonic 

AC AD AC AD 

pairs, then (ABCD) = - 1 ; i.e., -^ : ^ = ~ x > or BC^BD = °* 

AC AD 

But ^7-3 and -5-^; are the roots of (2); hence ABCD form 

v 
a harmonic group if the sum of the roots of (2); i.e., 2 — = 0. 

U 2 

As u 2 is not supposed to be on U, this is only possible if ^ = c; i.e., if 

(5) x 2 (ax x + by x + d) + ^(^ + cv x + e) + dx x +ey x + } = o. 



120 PROJECTIVE GEOMETRY. 

If four points A BCD, of which C and D are on Z7, are collinear 
and form a harmonic group, then the coordinates of A and B are 
related by (5). Keeping the point A fixed and letting B under 
condition (5) vary, it is plain that all points B restrained by these 
conditions lie on the straight line 

(6) x(ax x + by x +d) + y(bx 1 + cy x + e) + dx x + ey x + / = o. 

This line is called the polar 0} the point A (x x , y x ) with respect 
to U. 

Similarly, the polar of B(x 2 , y 2 ) is given by 

(7) x(ax 2 +by 2 + d) + y(bx 2 + cy 2 + e) + dx 2 + ey 2 + f = 0. 

If (x 2 , y 2 ) lies on the polar (6) of (x x , y x ), then (5) holds. But 
this can also be written as 

(8) x 1 (ax 2 +by 2 +d) + y 1 (bx 2 +cy 2 +e) + dx 2 +ey 2 +f = o, 

which is the condition that (x x , y x ) lies on the polar of (x 2 , y 2 ). 
Hence the theorem which has already been established before : 

// A is on the polar of B, then B is on the polar of A. 

From this it follows that the polars of the points of a straight 
line all pass through its pole, and, conversely, the poles of all rays 
through a fixed point lie on its polar. 

3. To establish the relation between the points of a straight 
line and the corresponding pencil of polars, assume first any four 
lines through a fixed point : 

p x = a x x+b x y+c x = o i 

p 2 = a 2 x+b 2 y + c 2 = o, 

p x -*p 2 = o, 

Cut these lines by any transversal and find the anharmonic 
ratio of the four points of intersection A X A 2 A 3 A 4 . For the sake of 
simplicity, choose the x-axis as this transversal, so that the distances 
of these points from the origin become 

£1 C 2 c \ *C 2 C x fiC 2 

~W "V ~b x -Xb 2 ' "h-fibl' 



THEORY QF CONICS. 121 

The anharmonic ratio is easily found: 

If now (x 1} y x )j (x 2 , y 2 ) are the coordinates of two points A u 
A 2 , then the coordinates of any point A 3 on A X A 2 are 

T^P "i^T' where '-4A- 

The polar of ^4 3 is 



I 


-/l 


#1- 


-;^ 2 


I- 


-;, 


*1- 


-Lr 2 



J^ +6 .2lz& +(/ 



- C -^p+« 



■-*=:?+/-* 



or, multiplying by i — ^ and rearranging, 

# (a^ + ^! + d) + y (&^ + c>'i + e ) + ^i + O'i + / 

- X \x(ax 2 + by 2 + d) + y(bx 2 +cy 2 + e) + dx 2 + ey 2 + }} =o. 

Designating the equations of the polars of A ly A 2 simply by 
Pi = °i p2 = °, the polar of A 3 will be represented by p x ~ Xp 2 = o. 
Analogously, the polar of a fourth point A 4 on A 1 A 2 is repre- 

A A 

sented by p 1 — P-p 2 = o, where fi= * * The anharmonic ratio of 

A 4 A 2 

. , . .11 A.o/i-1 /La/1i /li/Lo /l /±o A 

A., A 2 , A 3 , A 4 is evidently t~a : aa = ~aa : 1a = ~' Accord - 

ing to (9) the same is true of the four points of intersection of any 
transversal with the polars p 1 = o i p 2 = o, p 1 — ^p 2 = o, p x — fJ-p 2 = o 
of the points A ly A 2 , A 3i A 4 . Hence the theorem: 



122 



PROJECTIVE GEOMETRY. 



The range of points of a straight line and the corresponding 
pencil oj polars are projective. 

This follows also by considering the polar involution around 
a point and the corresponding involution of poles on its polar, 
as shown elsewhere. 

4. Equation or a Conic in Line-coordinates. 

To find the equation in line- coordinates u, v of a conic with 
the Cartesian equation 

(10) ax 2 +2bxy+cy 2 +2dx+2ey+} = o, 
consider the equation of a tangent 

(11) (ax 1 +by 1 +d)x+ (bx^cy^ e)y+ (dx t + ey 1 +f)=o 

at a point (x lf y ± ) of this conic. The line coordinates of this 
tangent are 

ax^-\-by t -{-d 



(12) 



v = 



dx ± + ey±+ J ' 
bx t -\- cy t -{- e 
dx^ey^f 



Conversely, the Cartesian coordinates x ly y ± expressed in terms of 
the line-coordinates u and v of the tangent at this point are, from 

(12), 

(cf— e 2 )u+ (de- bf)v+ (be- cd) 



(13) 






y x = 



(be— cd)u-{- (bd— ae)v-\- (ac— b 2 ) 
(de- bf)u+ (af- d 2 )v+ (bd- ae) 
(be— cd)u-\- (bd— ae)v+ (ac— b 2 ) 



But these values of x t and y 1 satisfy (10). Thus, substituting 

(13) in (10), we get the relation which exists between the line- 
coordinates u and v of the tangents of the conic (10), or the equa- 
tion of the conic in line-coordinates. After reduction this equa- 
tion becomes 

(14) (cf- e 2 )u 2 +2(de- bf)uv+ (af- d 2 )v 2 + 

2 (be— cd)u+ 2(bd— ae)v+ ac— b 2 =o. 



THEORY OF CONICS. 123 

Ex. i. Find the line-equations of 

x 2 y 2 
~y> it 79 1 j 



y 2 = px\ 
(x-a) 2 +(y-b) 2 -r 2 = o. 



x 2 +y 2 = r 2 \ 



Ex. 2. Establish the equation of a point of (14). 
Ex. 3. From the line-equation of a conic, 

au 2j r 2buv+ cv 2j r 2du-\- 2ev+ / = o, 

establish the Cartesian equation. 

§ 37. Theory of Reciprocal Polars. 

1. We have already discussed the principle of duality, § 22, 
in an elementary manner. In this section it will be seen that 
the principle follows directly from the theory of polars. 

To every point as a pole corresponds a straight line as a polar, 
and conversely. To two projective pencils producing a conic 
correspond, according to the theorem at the end of § 36, 3, two 
projective point-ranges which produce a conic as an envelope; 
i.e., to the points of a conic correspond the tangents of another 
conic, called the reciprocal of the first. In general, to any figure 
consisting of points and straight lines corresponds a figure con- 
sisting of straight lines and points, the polars and poles of the 
points and lines of the first figure. The anharmonic ratios of 
corresponding elements are the same in the original and reciprocal 
figure. 

The transformation thus established is called polar reci- 
procity and is contained in the slightly more general principle 
of duality. The polar of a point (x lf y t ) with respect to the 
conic U is given by 

(1) x(ax 1 +by 1 +d) + y(bx 1 +cy 1 + e) + dx 1 +ey 1 +f = o, 



124 PROJECTIVE GEOMETRY. 

or, introducing the line-coordinates, 

ax x +by x +d 

bx^+cy^+e 

, ~ dXi+eyt + f 

» 

(3) ocu+yv+i=o. 



Formulas (2) are the analytical expression for this trans- 
formation. To the point (x lt y ± ) corresponds the straight line 
with the coordinates (u, v). As the transformation is involu- 
toric; i.e., that the coincidence of a point and straight line neces- 
sitates the coincidence of their polar and pole, we can inter- 
change (x, y) with (x ly v x ), as has been already established. 

Designate now the original conic by U, the conic to be recip- 
rocated by K x and the reciprocal conic by K 2 . Assume U and 
K x as central conies. If the center of U is outside of K x , two 
tangents from it may be drawn to K x , which when reciprocated 
are two points of K 2 . As these tangents pass through the cen- 
ter of U, their poles will be infinitely distant. From this it fol- 
lows that K 2 is an hyperbola. If K t passes through the center 
of U, then only one real tangent can be drawn to K t at this point; 
i.e., K 2 will have only one infinite point (tangent) and is there- 
fore a parabola. When the center of U is inside of K v no real 
tangents from it can be drawn to K^ i.e., K 2 has no infinite 
points and is consequently an ellipse. Hence the theorem : 

According as the center of the original conic U is outside, on, 
or inside oj the conic K t to be reciprocated, the reciprocal conic K 2 
will be an hyperbola, a parabola, or an ellipse. 

2. According to (2) (x lf y t ) is the pole of the line with the 
coordinates (u, v). Suppose now that this line envelopes a 
circle of radius r and having its center in the origin of coordinates. 
The line-equation of this circle is 



(4) u*+v*= l 7 . 



THEORY OF CONICS. 125 

Taking U=(x— a) 2 + (y-ft) 2 - p 2 = o; i.e., a=i, b = o, e=i, 
d=—a, e=—ft, / = a 2 +/? 2 — p 2 , substituting these values in (2), 
and finally substituting the values of u and v, thus obtained in 
(4), we get 

*!-<* } \ ( yi-P 

or, expanded and rearranged, 

(5) ( y2 ~ « 2 )^i 2 — 2a(3x 1 y 1 + (r 2 — /5 2 )v! 2 + 2a:(a: 2 +/5 2 — p 2 — r 2 )x 1 

+ 2p(a 2 +p 2 -p 2 -r 2 )y 1 +r 2 (a 2 +p 2 )-(a 2 +p 2 -p 2 ) 2 = o. 

This is the equation to which the poles of all tangents of 
(4) are subjected. Hence (5) is the equation of the conic recip- 
rocal to the circle ^ 2 +v 2 = r 2 with respect to the circle (x— a) 2 
+ (y—P) 2 = p 2 - Here the characteristic determinant of (5) is 

Evidently r>, =, <o, according as a 2 +/9 2 >, =, >r 2 . 
Hence, according as the center of U is outside of, on, or inside 
of (4), the reciprocal conic (5) is an hyperbola, a parabola, or 
an ellipse, which is in agreement with the previous result. 

Ex. 1. What is the reciprocal of a polygon circumscribed 
to a conic with respect to this conic? 

Ex. 2. Find the reciprocal of the point Au+Bv+C = o. 

Ex. 3. Find the reciprocal of the envelopes u 2 — v 2 = o; 
u 2 —v 2 =i; uv = u 2 +v 2 . 

Ex. 4. Discuss reciprocation in the case 'where in formulas 
(2) the determinant 



a b d 
bee 
d e j 



= 0. 



126 PROJECTIVE GEOMETRY. 

Ex. 5. Given the polar-reciprocal transformation 

Ax+By+D 
U ~Dx+Ey+F 7 

Bx+Cy+E 



v = 



Dx+Ey+F' 



by which to every point (x, y) corresponds a straight line (w, v), 
and conversely. 

Establish the equation of the conic, for which every point 
(x, y) coincides with the corresponding line (u } v). 



§ 38. General Reciprocal Transformation. Polar Systems. 

1. Formulas (2) of the foregoing section may be generalized 
by setting 

a 1 x-\-b 1 y+c 1 



(1) 



u = 



v = 



aapc+b 3 y+c 3 ' 

a 2 x+b 2 y+c 2 

a 3 x+b 3 y+c 3 * 



(2) 



Solving (1) for x and y, we get 

(b 2 c 3 - b 3 c 2 )u+ (b 3 c t - \c^v-\- (b^- b 2 c t ) 
{a 2 b 3 — a 3 b 2 )u+ {a 3 b 1 — aj) 3 )v+ (a 1 b 2 — ajbj)' 
(a 3 c 2 — a 2 c 3 )u+ {a x c 3 — a 3 c t )v+ (p,£ x — a t c 2 ) 



x = 



y= 



(a 2 b 3 - a 3 b 2 )u+ (a 3 b t — a l b 3 )v+ (aj) 2 — a 2 b t y 



To a straight line 



ax+by+c = o 



corresponds by this transformation a point with the line-equation 



(3) 



' WVa- V2) + K a $T- a 2 c s) + c(a 2 b 3 - a 3 b 2 ) } u+ 
WVi- V 3 ) + K a i c 3~ Wi) + cia^- aj) 3 ) }v+ 
I 1 0(6^3— Vi) + K a £i- <V 2 ) + c(aj) 2 - ajb t ) } =0. 



THEORY OF CONICS. 1 27 

Conversely, to a point 

au+bv-\-c = o 
corresponds a line with the equation 
(4) (0^1+ ba 2 + ca 3 )x-\- (ab 1 +bb 2 + cb 3 )y+ (ac ± +bc 2 + cc 3 ) =0. 

If four points, of which no three are collinear, with the equa- 
tions a i u+b i v-\-c i =o (* = i., 2, 3, 4) are given; and also four 
arbitrary lines, of which no three are concurrent, with the equa- 
tions a i x+p i y+r i =o (* = i, 2, 3, 4), we can let these points and 
lines correspond to each other in a reciprocal transformation 
by setting 

f(a i a 1 +b i a 2 + c l a 3 ) — a*(a*c t + &*c 2 + c*c 3 ) =0, 

*=i| 2 > 3, 4, 
^'(afy + 6*6 2 + c%) - pXa\ + &*c 2 + c%) = o. 

These are eight equations with the eight unknown ratios 

a. a 2 a 3 b. b 2 b 3 c t c 2 1.1,1 , 

— , — , — , — , — , —,—,—, from which the latter may be found 
c 3 c 3 c 3 c 3 c 3 c 3 c 3 c 3 

definitely. Hence the theorem: A quadrilateral and a quad- 
rangle always determine a reciprocal transformation in which 
they correspond to each other. The reciprocal transformation is 
the most general dualistic transformation and includes polar 
reciprocity as a special case. This is easily recognized by 
comparing formulas (2) of § 37 and (1) of this section. 

2. We shall now determine those lines of the coincident planes 
(u, v) and (x, y) which coincide with their corresponding points. 
A line with the coordinates u and v passes through the point with 
the coordinates x and y if 

Ux+vy+i=FO. 



128 PROJECTIVE GEOMETRY. 

Hence the points (x, y) whose corresponding lines (u, v) accord- 
ing to (i) pass through them satisfy the condition 

a.x-\-b.y-\-c< a 9 x+b?y+c~ 
1 V V + 2 ' 2 % -f i =o, 

^ 3 ^+ ^+ ^ ^ 3 ^+ t> 3 y+ H 

or 

(5) ^* 2 + (b x + a 2 )xy+b 2 y 2 + (c t + a 3 )x+ (c 2 +b 3 )y+c 3 =o, 

which represents a real or imaginary conic C. Conversely, for the 
lines (u, v) whose corresponding points (x, y) lie on them, we have 
the condition 

(6) (b 2 c 3 - b 3 c 2 )u 2 + (b 3 c x - b x c 3 + a 3 c 2 - a 2 c 3 )uv+ (a x c 3 - a 3 Cj)v 2 + 

(K c 3- Vi+ ajba— <h/> 2 ) u + ip-f\— a i c 2+ <hh— aj> 3 )v+ 

(a l b 2 —a 2 b 1 )=o,. 

which represents a conic of the second class r. The conies C 
and r are generally different, as may be seen by applying the 
results of § 36, 4, to equations (5) and (6). 

To every point of C correspond the two tangents from it to 
r\ conversely, to every tangent of r correspond its two points of 
intersection with C. If C and r have a point P in common, then 
to P on C correspond two coincident tangents to r at P, so that 
their corresponding points also coincide at P. This is only pos- 
sible when C and P are tangent at P. From this it follows that 
the two conies C and r are doubly tangent, and as there is no dis- 
tinction analytically, we may say that in case of no real intersec- 
tions the conies C and r have two imaginary tangencies. 

3. From (3) it is seen that to a pencil 

ax + b y + c + X (a'x + b' y + c f ) = o 

corresponds a range ; and the vertex of the pencil corresponds to 
the line of the range. The converse (apply (4)) is also true. Let 
now U and S be the points of tangency of C and r, and u and s 
the tangents at U and S, and T their point of intersection, and 
consider the planes of (u, v) and (x, y) as made up of lines and 
points and points and lines respectively; i.e., to a couple (w, v), 



THEORY OF CONICS. 129 

(au+bv+ 1=0) in one plane corresponds dualistically a couple 
(x, y), (ax+[3y+ 1=0) in the other plane, and conversely. 

No matter whether we consider T as belonging to one or the 
other plane, SU is the corresponding line in both cases. In the 
reciprocal transformation T and SU are therefore in the relation 
of involution. For both conies SU is the polar of T. 

The question is whether it is possible to find a reciprocal trans- 
formation for which the involutoric property is true in general. 
For this purpose consider a point 

au+bv+c = o 
in the plane (u, v) and the same point ( — ),(—) in the plane (x, y). 



c r \c 

To the point ( - , - ) in the x^-plane corresponds the line with 
the coordinates 

aa t + bb ± + cc t aa 2 + bb 2 + cc 2 . 

(7) u== , 1.1. , — > v = TTT - ; — in the wv-plane. 

w/ aa 3 +bb 3 +cc 3 aa 3 +bb 3 +cc 3 r 

To the point (au-\-bv+c=o) in the w^-plane corresponds the 
line 

(aa 1 +ba 2 +ca 3 )x+ (ab l +bb 2 +cb 3 )y+ (ac x -{-bc 2 -\-cc 3 ) = 

in the x^-plane. Its line-coordinates are 

aa< + ba, + ca» , ab. + bb 2 + cb» 
v y ac x -\-bc 2 -\- cc 3 ac x -\- bc 2 -\- cc 3 

For an involutoric relation the two lines (7) and (8) must be 
identical. This will be the case when b x = a 2 , c 1 = a 3 , c 2 = b 3 \ i.e., 
if the transformation (1) has the form 

a 1 x+b 1 y+c 1 b t x+b 2 y+c 2 

(9) U = ; ; , V = ; ; . 

Kyj c x x+c 2 y+c 3 ' c 1 x+c 2 y+c 3 

According to (2), § 37, these are the formulas for a transforma- 



*3° 



PROJECTIVE GEOMETRY. 



tion by reciprocal polars. To prove this directly the equation (5) 
of the conic C now becomes 



(10) 



a x x 2 + 2b 1 xy+ b 2 y 2 + 2C 1 x J r 2C 2 y+ c 3 = o. 



To a point ux^vy^ 1=0 now corresponds the line (accord- 
ing to (4)) 

(11) (a 1 x 1 +b 1 y 1 + c t )x+ (Mi+ b 2 y^ c 2 )y+ (c 1 x 1 +c 2 y 1 +c 3 )=o. 

This, however, is the polar of the point (x 19 y x ) with respect to 

(10). 

An involutoric recipcocal transformation is therefore a trans- 
formation by reciprocal polars. 

In this case the conies C and T coincide. 

4. The line- coordinates given in (7) and (8) are also identical 
if corresponding numerators and denominators are proportional. 
Designating the proportionality factor by A, these conditions 
assume the form 



(12) 



a^i — X)a+ (b t — Xa 2 )b -\-(c 1 —Xa 3 )c=o, 
(a 2 -Xb 1 )a+ b 2 (i-X)b + (c 2 — Xb 3 )c=o, 
(a 3 — Xc 1 )a+ (b 3 — Xc 2 )b +c 3 (i — X)c =0; 



but consistency of these equations requires the vanishing of their 
determinant, or 



(13) 



^(1 — A) b 1 —Xa 2 c 1 —Xa 3 
a 2 —Xb 1 b 2 (i — X) c 2 — Xb 3 
a 3 — Xc x b 3 — Xc 2 c 3 (i — X) 



o. 



This is the case, first when b± = a 2 , c 1 = a 3 , c 2 = b 3 , and A=i, as 
discussed under 3; secondly, by expanding the determinant 
according to ascending powers of X and solving the cubic in X. 
Thus three values for X are obtained which make the determinant 
vanish. One of these values is always real, so that there is at 
least one real line which with its corresponding point forms an 
involutoric couple. 



THEORY OF CONICS. 131 

To push the investigation of involutoric reciprocity one step 
further we may put the condition for the equality of u and u' y 
and v and v', in (7) and (8) in the form 

(14) (a 1 x-{-a 2 y-{-a 3 )(a B x+b 3 y-hc a )- (a 1 x+b 1 y+c 1 )(c 1 x+c 2 y+c 3 )=o 1 

(15) (b ± x+ b 2 y+ b 3 ) (a 3 x+ b 3 y+ c 3 ) - (a 2 x+ b 2 y+ c 2 ) (c x x+ c 2 y+ c 3 ) = o, 

where x= — , y = — are the coordinates of the original point. 

There are generally four solutions of (x, y) which satisfy 
(14) and (15) simultaneously. Of these, one is the point of inter- 
section of the lines a 3 x+b 3 y+c 3 = o and c 1 x+c 2 y+c 3 =o, which, 
however, is to be excluded. In fact, according as this point is 
considered as belonging to one or the other plane, (u, v) or (V, v'), 

u a 1 x-\-b 1 y-\-c 1 
the lines through the origin with the slopes — = — — -7 — — — > 

v a 2 x-\- o 2 y-\- c 2 

u' a x x+a 2 y+a 3 , . . , 

-7=7 — —7 — —7- correspond to it. Hence, as we have found 
7/ b x x+b 2 y+b 3 ^ ' 

before, there are in general only three involutoric pairs in a 

reciprocal transformation; they are determined by the three 

remaining points of intersection of (14) and (15) and form a 

triangle UST, according to 3, in which TV and TS correspond 

to the points U and 5, and T to the line US, involutorically. 

Hence in a reciprocal transformation there is generally only 

one involutoric pair (T 1 , US) which is not coincident. 

Suppose that this be true for a second pair of this kind, then 
(14) and (15) would have a fifth common solution which is only 
possible when the two are identical. Hence the theorem: 

// a reciprocal transformation contains two non-coincident 
involutoric pairs, then all its pairs are involutoric; the transforma- 
tion is a so-called polar reciprocity. 

By two non-coincident involutoric pairs the polar reciprocity is 
fully determined. 

To prove this last theorem equations (9) and the equation for 

u 

— obtained from them may be written in the form 



132 PROJECTIVE GEOMETRY. 

a i ^1 c i , N C 2 

-x+ — y+-(i-ux)--uy-u=o, 

Cg C3 (,3 t/3 

^1 ^2 ^1 C 2 , 

— #+ — y- vx+— (1— ^3;)— v=o, 

c 3 c 3 c 3 c 3 

a i W , n ^2 C l C 2 

—vx+ —(vy— ux) — —uy-\- —v u = o. 

c 3 c 3 C 3 ' c 3 c 3 

Giving {x, y) two arbitrary values and (u, v) correspondingly 

two arbitrary values, six equations with the only five unknown 

. . a t b t b 2 c ± c 2 
quantities — , — , — , — , — , are obtained. Designating the two 
£3 ^3 ^3 ^3 ^3 

pairs by (x ly yj, (x 2y y 2 ) and (u lt v t ), (u 2 , v 2 ), the determinant of 
the six equations becomes 



= 



In fact multiplying the six rows successively by v v — v 2 , — u lf u 2 , 
+ 1, —1, as indicated, after this multiplication, the sum of the 
first four rows is equal to the sum of the last two, which shows 
that any of the six equations may be expressed in terms of the 
five remaining ones. The above five quantities are therefore 
uniquely determined, which proves the theorem. 

5. In a polar reciprocity, or simply in a polar system, two 
pairs A, a and P, p determine at once a third. Indeed the line 
c joining A and P is the polar of the point of intersection C of a 
and p. The pole of AC is the intersection of a and c, say B. 
Thus, starting .with two pairs, we have constructed a triangle 
ABC, whose vertices are the poles of its opposite sides. Such 
a triangle is called a self- polar triangle (§ 14). Clearly in every 
polar system there are an infinite number of self-polar triangles; 
but by such a triangle a polar system is not completely determined. 



*1 


x x 




ft 


(i-*W 





-U x y x 


— u x 


^2 


x 2 




y 2 


(l-U 2 X 2 ) 





-u 2 y 2 


-u 2 


u x 







X-% 


-V& 


y± 


(1-^1) 


-Vi 


u 2 







x 2 


-v 2 x 2 


y* 


(i-v 2 y 2 ) 


-v 2 


I 


v x x t 


(*w 


l~ ^1^1) 


v l 


-u.y, 


— u x 





I 


v 2 x 2 


(v 2 y 


2 ^2^2/ 


v 2 


-1W2 


-u 2 






THEORY OF CONICS. 133 

To do this, another pair, like P, p, must be added to the given 
triangle. 

6. Without following the subject of polar systems further we 
remark that the great geometer von Staudt has made it the back- 
bone of his geometry of position. In this connection conies 
appear as special properties of polar systems and no distinction, 
or separate treatment of real and imaginary elements, is necessary. 

In view of the various methods applied in this work, we have 
found it advisable to be satisfied with the foregoing short account. 

It would be very valuable if some geometer could show how, 
with polarity as a base, projective geometry might be made as 
simple and as accessible to the applications as the traditional 
methods. 

§ 39. Theorems of Pascal and Brianchon. 1 

1. Assume six points A, B, C, D, E, F in any order on a 
conic and consider any two of them, say A and C, as vertices 
of pencils of rays in the conic, Fig. 49. Then 

(A -BCDEF) = C -BCDEF). 

Cutting these pencils by the lines ED and EF respectively, 
two projective point- ranges, 

{B X C 1 D,E 1 F 1 ) = {B 2 C 2 D 2 E 2 F 2 ), 

are obtained, and as E 1 is identical with E 2 it follows (§9) that 
the two ranges are perspective. Hence B ± B 2 , C X C 2 , D ± D 2 , E X E 2 , 
F 1 F Z are concurrent at a point B 3 . This will be true no matter 
how the six points may be distributed over the conic, pro- 
vided the foregoing order of the points is followed. The lines 
followed in the order A BCDEF form now a closed hexagon, 

1 Pascal (162 3-1 662) discovered his theorem when sixteen years of age and 
called it Hexagramma Mysticum. It appeared first in Pascal's "Conic Sections," 
which was published in 1640. 

Brianchon (i 785-1 864) published his theorem in 1806 in the Journal de 
l'Ecole Polytechnique, Vol. XIII. 



134 



PROJECTIVE GEOMETRY. 



and we may call opposite sides of this hexagon lines which are 
separated by two adjacent sides of the hexagon, which is all in 
analogy with the regular hexagon. The pairs of opposite sides 
intersect at B t , B v B 3 , three collinear points. As the six points 




Fig. 49. 

were arbitrarily selected, this is generally true, hence Pascal's 
Theorem : 

In any hexagon which is inscribed in a conic, the three pairs 
of opposite sides intersect in three collinear points. 

We shall call such a line of collinearity a Pascal line. 

By reciprocation, Fig. 50 (§ 37), we obtain immediately in 
its generality Brianchon's Theorem: 




Fig. 50. 

In any hexagon which is circumscribed about a conic, the 
three principal diagonals are concurrent. 



THEORY OF CONICS. 135 

We shall call such a point of concurrence a Brianchon point. 

2. Salmon, in his treatise on Conic Sections, 1848, gave a 
remarkably simple proof for Pascal's theorem, based upon the 
abbreviated designation of straight lines in analytic geometry. 
Let A=o, 5 = o, C=o, D=o, E=o, F=o be the equations of 
the sides of any hexagon inscribed to a conic, and G = o the equa- 
tion of the straight line joining the vertices (A =0, F = o) and 
(C = o, D=o). Then 

A-C-XB-G = o, F-D-/xEG = o 

are two forms in which the equation of the given conic may be 
written. From these two forms we get 

A-C-F-D^G(W-fxE). 

Now the points (A—o, D=o) and (C = o, F = o) are not 
situated on the line G=o, consequently they must lie on the 
line IB— jiE=o. In other words, the points {A =0, D=o), 
(C=o, F = o), (B = o, E = o) are collinear, and as they are the 
points of intersection of pairs of opposite sides in the hexagon, 
Pascal's theorem is proved. 




Fig. 51. 

By considering A =0, etc., as the line-equations of the six 
vertices of a hexagon circumscribed to a conic, Brianchon *s 
theorem may be deduced in a similar manner. 



136 PROJECTIVE GEOMETRY. 

3. Assuming as a conic a degenerate hyperbola, consisting 
of two intersecting lines and on each three points, say A, E, C 
and D, B, F, Pascal's theorem still holds; i.e., the points B lt 
B 2 , B 3 are collinear. 

If AB \\DE and EF \\ CB, then B x and B 2 and consequently 
also B 3 are infinitely distant; i.e., also CD \\AF, Fig. 51. Hence 
the special theorem: 

// on each 0} two intersecting lines three points A, C, E and 
B, D, F are chosen, so that AB is parallel to DE and EF parallel 
to BC, then CD is also parallel to AF. 

In this special form Hilbert in his Foundations of Geometry, 1 
p. 28, uses Pascal's theorem to establish a non-Archimedean 
geometry. 

Ex. 1. Prove Pascal's special theorem directly. 

Ex. 2. Establish the dualistic of Ex. 1. 

Ex.3. If 4=o, rA + bB = o, r B' + aA' = o, A' = o, aA'+ r 'B 
= 0, /A + bB' = o (where a, b, y, f are numerical factors and 
A, B, A f , B r linear expressions in x and y) are the sides of a hex- 
agon, prove that this hexagon is inscribed to the conic 



and that 



aAA'-bBB' = o, 

yfA — abA' = ° 



is the Pascal line of the hexagon. (Bobillier, 1828.) 

Ex. 4. If six points on a conic are given, it is possible to pass 
in five different ways from any point to the others. From each 
of these four different paths, not chosen before, may be taken to 
join the remaining points; from each of these' three different 
paths may be selected; and so forth. Finally the original point 
is reached in 5'4'3-2-i = i2o ways; but as each closing side is 
contained in one of the original paths, it is evident that only 
120:2 = 60 different closed hexagons can be formed. Hence 
with six points on a conic may be formed sixty different hexagons 
and consequently sixty different Pascal lines. 

1 Grundlagen der Geometrie, Teubner, Leipzig, 1899. 



THEORY OF CONICS. 137 

Between these lines exist a number of interesting relations. 1 

Verify the following propositions in a regular hexagon: 

The sixty Pascal lines intersect each other three by three in 
twenty points G (Steinerian points). (Steiner's theorem.) 

Besides these points G, the sixty Pascal lines have, three by 
three, sixty other points H in common. (Kirkmann's theorem.) 

There are twenty lines g each of which contains a point G 
and two points H. Four by jour of these lines pass through 
fifteen points J. (Cayley's theorem.) 

The points G lie jour by jour in fifteen straight lines J. 
(Steiner's theorem.) 

Designating the original six points by 123456, then a Steiner- 
ian point is given by the intersection of the Pascal lines of the 
three hexagons 123456, 143652, 163254. 

Ex. 5. State the dualistic of the foregoing theorems. 



§ 40. Applications of PascaPs and Brianchon's Theorems. 

1. Construction oj a conic when five of its points are given. 

The practical importance of Pascal's and Brianchon's theorem 
lies in the possibility of constructing an unlimited number of 
points and tangents of a conic, when five of its determining ele- 
ments are given. 

Let ABODE be five points of a conic and AB, BC, CD, DE 
four consecutive sides of an inscribed hexagon. In Fig. 52, it 
is clear that the Pascal line p passes through B ly the point of in- 
tersection of AB and DE. Now there are an infinite number of 
points F possible on the conic and consequently an infinite num- 
ber of Pascal lines through B v Thus to every point F on the. 
conic corresponds one Pascal line through B v Hence, assuming 
any line p through B x , the line EF passes through the intersection 
B 3 of BC and p. In a similar manner the line FA is obtained 
by joining A with the point of intersection B 2 of CD with p. 

1 See Salmon-Fiedler, Analytische Geometrie der Kegelschnitte, Vol. II, pp. 
459-466, 5 th edition. 



I 3 8 PROJECTIVE GEOMETRY. 

The point where the produced lines of EB 3 and AB 2 meet is 
evidently the required point F on the conic, corresponding to the 



chosen Pascal line p. Repeating the same construction for every 
line p through B ly all points of the conic are obtained. 

To construct the tangent at any point of the conic, say A, 
consider F infinitely close to A. Apply the general construction 
of p for ABCDEF, then the line joining B 2 with A is the tangent 
at this point. 

2. Construction of a conic when five of its tangents are given. 

Let a, b, c, d, e be the given tangents, Fig. 53, forming five con- 
secutive sides of a circumscribed hexagon. The line b 1 joining 
the points of intersection of a and b, and d and e, passes through 
the Brianchon point P. Now, every sixth tangent determines 
another point P on b v Conversely, every point ? on ^ deter- 
mines a sixth tangent of the conic. Thus, to find a sixth tangent 
/, assume any point on b x as the Brianchon point P. Then the 
line through be and P will be the line b 2 cutting e in the point 
where also / cuts. In a similar manner, the line joining the point 
of intersection of c and d with P is b 3 , which, when produced, cuts 
a in the same point as /. Hence the line joining the points of 
intersection of e and b 2 , and b 3 and a, is the sixth tangent corre- 
sponding to the chosen P. Repeating this construction for all 
points of b lf all tangents of the conic are obtained. 



THEORY OF CONICS. 139 

To construct the point of tangency of any tangent, we may 
consider this one as two coincident tangents (consecutive), say a 
and /, and these with the remaining four, when subject to the 
general construction of the Brianchon point, lead to the required 
point of tangency. 

3. By the same methods conies may also be constructed when 
they are determined by mixed elements; i.e., points and tangents, 
always five in number. In these problems a tangent appears as 



Fig. 53. 

a line joining two consecutive points, and a point as the point of 
intersection of two consecutive tangents. The same construc- 
tions may also be extended to cases where one point, two points, 
or one tangent is infinitely distant. 

Ex. 1. Given five points of a conic; to construct the tangents 
at these points. 

Ex. 2. The dualistic of Ex. 1. 

Ex. 3. Given three points and the directions of the asymp- 
totes of an hyperbola; to construct any number of points of the 
hyperbola. 

Ex. 4. Given four tangents of a parabola (one tangent is 
infinitely distant). To construct any number of its points. 

Ex. 5. Given four points and a tangent of a conic; construct 
other points of the conic. 



140 PROJECTIVE GEOMETRY. 

Ex. 6. Dualistic of Ex. 5. 

Ex. 7. Given three points and two tangents of a conic. To 
construct it. Also make the dualistic construction. 

Ex. 8. Given three points, a tangent, and its point of tan- 
gency; construct the conic. 

Ex. 9. Given the two tangents at the given vertices of an 
ellipse or hyperbola and a third tangent; to construct any 
number of tangents. 

Ex. 10. Given the two asymptotes and a tangent of an hyper- 
bola; to construct it. 

Ex. 11. Given the axis, vertex, and two other points of a 
parabola; construct it. 

Ex. 12. Given three points and an asymptote of an hyper- 
bola; to construct it. 



§ 41. Conies in Mechanical Drawing and Perspective. 

1. To inscribe an ellipse in a parallelogram. 

The middle points of the sides shall be the points of tangency 
of the ellipse. Two points of tangency may be designated by A B 
and CD, and the third by E, Fig. 54. The explanation of the con- 
struction of points of the ellipse by Pascal's theorem is identical 
with that of Fig. 52, § 40, and is apparent from Fig. 54. By 
assuming a second Pascal line through L with points H and / 
corresponding to M and N on the first Pascal line, a second 
point G is obtained. The same construction repeated for other 
Pascal lines through L gives further points of the ellipse, so 
that the ellipse through these points may be sketched free-hand 
or by mans of a curved ruler. In this figure the ellipse appears 
manifestly as the product of two projective pencils with A and 
E as vertices. In fact, 

(AMCH) = (KNCJ), 

since these points are projected by one and the same pencil 
through L. Taking a Pascal line parallel to KC and desig- 



THEORY OF CON1CS 



141 



nating its point of intersection with AC produced by J, then 
the point corresponding to J has moved to infinity on KC, and 




Fig. 54. 

to the Pascal line LI corresponds the point E on the ellipse. 

Now 

(AMCI) = (KNC 00), 

and as AC =CI, these ratios become 

1 AM KN 
2CM~CN' 



But there is also 



(AMCI) = (OF x C<x) 



hence 



KN OF, 

cn~cf; 



142 



PROJECTIVE GEOMETRY. 



From this it follows that the rays AN and EM of the pencils 
through A and E trace on KC and OC similar point-ranges. 
If, therefore, KC and OC are divided into any number of equal 
parts and the division-points are numbered from K to C and 
from O to C, then the rays joining E and A with equal numbers 
on KC and OC intersect each other in points of the ellipse. In 
a similar way this construction may be extended to the remain- 
ing three quarters of the ellipse. The same method may obviously 
be applied to rectangles and squares. See Figs. 37, 38; § 27. 

2. To inscribe an ellipse to any quadrilateral. 

A quadrilateral may be considered as the perspective of a 
square, and it must therefore be possible to apply the previous 
construction to any quadrilateral. The distances KC and OC, 
Fig. 55, must now be divided perspectively into a number of 
equal parts. The fundamental principle of perspective division 
is the following: 

// KC as a side of a rectangle AKCO, in perspective, shall be 
divided into two equal parts, draw the diagonals AC and KO and 
join their point of intersection W to the point X, where AK and OC 
produced meet. WX cuts KC at its middle point, M. 




Fig. 55. 



In the first place, the points AB, C, E, etc., were obtained by 
the application of this principle to the given quadrilateral. 

By the same principle KM and CM may be again bisected. 



THEORY OF CONICS. 143 

OC may be subdivided in the same manner. Fig. 38, § 27, illus- 
trates the construction of an ellipse inscribed in a quadrilateral 
by this principle. 

The problem to inscribe an ellipse into a quadrilateral appears 
in a great number of special forms in perspective. For example, 
a trapezoid may be considered as the perspective of a square 
having two opposite sides parallel to the picture-plane, as in 
the case of window- frames and doors. 

3. To construct a parabola having the vertex, the major axis, 
and a point given. 

Let, in Fig. 56, the vertex be designated by AB, the infinitely 
distant point of the axis by DE, and the third point by C. Evi- 
dently any line p parallel to the tangent at the vertex may be 
considered as a Pascal line. The construction 

AB ) BC) CD ) 

DE \ L > EF\ ' FA p 

for the assumed Pascal line p gives us a point F of the parabola. 
If p varies, the point-ranges traced by M, N, F x on AC, KC, AK 
are all similar. Hence, dividing KC and AK in any number 
of equal parts, numbering the division-points from K to C and 
from A to K, a line joining A to any number on KC and a line 
through the equal number on AK, parallel to the axis, cut each 
other in a point of the parabola. 

4. Construction of a parabola which is the funicular polygon of 
a uniformly distributed load on a horizontal beam. 

If a load is uniformly distributed on a horizontal beam, then 
the funicular polygon is a parabola limited by points in perpen- 
diculars through the extremities of the beam. The tangents at 
the extremities of the parabola are known; they are parallel to 
the extreme lines of the force polygon. Designate in Fig. 57 
the tangents at the extremities by ab and de, and the infinite 
tangent by c. Then the Brianchon points P are situated on ad, 
and lines through P parallel to ab and de (the tangents at the 
extremities) cut these in two points x and y through which the 



144 



PROJECTIVE GEOMETRY. 



sixth tangent / passes. If P moves on ad, then the points x and y 
trace on ab ■ Q and de • Q two equal point-ranges. Hence, dividing 




Fig. 56. 

these distances in a number of equal parts and numbering them 
from ab and Q, the lines joining equal numbers are tangents of 
the required parabola. To find the point of tangency of /, 
replace ab by c, de by /, / by ab, and the infinite tangent by de, 




de being the infinite point of the axis of the parabola. For this 
arrangement P is also the Brianchon point, and the construction 
shows at once that the required point of tangency T is cut out by 
a line through P parallel to the axis of the parabola. 



THEORY OF CONICS. 



*45 



To sum up, we have the following construction for a parabola 
touching the sides of an isosceles triangle ABC, AC = BC, at A 
and B: Divide AB, BC, and CA into the same number oj equal 
parts and number the division- points from A to B, from A to C, 
and from C to B. The lines joining equal numbers on AC and 
CB are tangents of the required parabola, and the perpendiculars 
from corresponding equal division- points on AB cut tfiese tangents 
in their points of tangency. 

5. Construction of an equilateral hyperbola when its asymptotes 
and the tangent at a vertex are given. 

In Fig. 58 designate the , asymptotes by ab and de, and the 
tangent at the vertex by c. Let c cut the asymptotes at A and B. 




Fig. 58. 

In this case the Brianchon points are infinitely distant. Hence, 
drawing through A and B two parallel lines in any direction, 
cutting the asymptotes at C and D, the line joining C with D 
will be a required tangent of the hyperbola. To study the metrical 
relations of this hyperbola, we have 

AAOD^ACOB, 



146 PROJECTIVE GEOMETRY. 

hence DO:AO = BO:CO, 

or CODO=AOBO = constant. 

Designating the distance of the vertex M from the asymptotes 
by k, there evidently is 

CO - DO = 4k 2 , 

a relation which holds for any tangent. Hence the triangle 
between the asymptotes and any tangent has a constant area. (This 
is true for any hyperbola, as might be proved in a similar manner.) 

To find the point of tangency of CD, replace the designation 
f by ab, ab by de, de by c, and AB by /. (The student should 
make a new figure.) Join the point of intersection Q of AB 
and CD to O and produce to the point of intersection R with 
BC\ then R is the new Brianchon point and the line through 
R parallel to OD cuts CD in the required point of tangency N. 
As A BCD is a quadrilateral in which AD || BC and O and Q 
are diagonal points, BR = RC, hence also DN = CN. The point 
of tangency bisects, therefore, the tangent between the asymptotes 
(general proposition). 

Designating the coordinates of N by x and y, we have x=^CO, 
y = \DO\ hence xy = \CO-DO, and as CO • DO = 4k 2 , 

xy = k 2 . 

This is the equation of the hyperbola referred to its asymptotes. 
A full treatment of this case was given in view of its importance 
in the graphical representation of Boyle's law expressing the 
relation of the volume x and the pressure y of a gas. 

§ 42. Special Constructions of Conies by Central Projection 
and Parallel Projection. 1 

1. Given five points 0} a conic, to construct a circle of which the 
given conic is a perspective. 

1 For the collection of these problems the author is indebted to Dr. Karl Doehle- 
mann's Geometiische Transformationen, I. Teil, Goschen, Leipzig, 1902. 



THEORY OF CONICS. 147 

In §§20 and 27 it has been shown analytically and synthetic- 
ally that every quadrilateral may be considered as the perspec- 
tive of a rectangle which is always inscribed to a certain circle. 
It is therefore possible to construct four points A, B, C, D of a 
rectangle as the points whose perspectives are four given points 
A' , B f , C , D f of the conic. As has been explained in § 27, the 
two diagonal points M' and N r determine the vanishing-line and 
are the vanishing-points of the pairs of parallel sides AB, CD and 
AD, BC. The center of perspective collineation is situated on a 
circle over M'N' as a diameter, and AB, CD and AD, BC are 
respectively parallel to SM' and SN', Fig. 59. The axis of col- 

q' m' ?' - N^ q/_ 






w 




Fig. 59. 

lineation s must be chosen parallel to q' or M'N 1 at any distance 
from it. From 5 and s, A BCD is perfectly determined. To 
determine its position in space the distance-circle with S as a 
center must be given. There are therefore three elements, S, s, 
and distance- circle, which determine A BCD, of which A'B'C'D' 
is a perspective, completely. Hence there are oc 3 rectangles in 
space of which A'B'C'D' is a perspective. If now E, of which 
E f is the perspective, shall also be situated on the circle through 



148 PROJECTIVE GEOMETRY. 

A BCD, notice that AC is a diameter, hence A EC a right angle. 
Consequently if we produce A'E' and C'E' to their intersections 
P r and Q' with q', the center 5 necessarily also lies on the circle 
over P'Q' as a diameter. 

In the figure the construction of ABCDE has been removed 
parallel to s in order to make it clearer. In this construction we 
may dispose arbitrarily of 5 and of the distance -circle. Hence 
there are go 2 circles in space which may be transformed into a 
given conic by perspective, under the given conditions. To make 
this proposition general it must be remembered that the analytic 
expression for perspective involves three essential parameters. If 
a translation of the center of perspective is added, two more con- 
ditions enter, so that, together with the choice of the distance - 
circle, six constants perfectly determine a central projection. If, 
therefore, the general equation 

ax 2 -\- 2bxy-\-cy 2 + 2dx+ 2ey J r f = o, 

by means of this projection, is transformed into 

Ax 2 + 2Bxy+ Cy 2 + 2Dx+ 2Ey+F = o, 

this equation contains those six constants. If this equation shall 
represent a circle, the conditions A=C, B = o must be satisfied, 
so that of the six constants only four remain independent. 

Hence the theorem: 

A conic may be considered as central projection of oo 4 circles in 
space. 

Ex. i. Given five tangents of a conic; to construct a circle of 
which the given conic is a perspective. 

Hint: Any four of the given tangents may be transformed 
into a rhomb circumscribed to the required circle. The diagonals 
of this rhomb are perpendicular and intersect at the center M of 
the required circle. Furthermore, the piece of the fifth tangent 
between two parallel sides of the rhomb appears under a right 
angle from M. 

Ex. 2. Any two conies in a plane may be considered as the cen- 
tral projection of two circles. (Monge.) 



THEORY OF COXICS. 149 

The two circles are supposed to be in one and the same plane. 
Now every point of the plane may be taken as the center of pro- 
jection, so that there are oo 3 -oo 3 =oo c central projections of a 
plane. Transforming the equations of the given conies, six 
parameters are introduced of which we may dispose arbitrarily. 
In order that the transformed equations represent circles, four 
conditions must be satisfied, so that there is still an infinite 
number of possibilities for the problem left. 

In case that the given conies have four real points of inter- 
section, imaginary elements are introduced in the solution. The 
validity of the geometrical problem in this case is maintained by 
Poncelet's principle of continuity. 1 

Ex. 3. Prove that any conic and a straight line in its plane 
may be projected centrally into a circle and the infinite line of its 
plane. 

2. Conies as intersections of right cones. 

Let in a plane perpendicular to the paper, Fig. 60, a conic K 
with the foci F, F x and the vertices A,A X be given. At one of the 
foci, say F, construct any sphere, S, tangent to the plane of the 
conic, and from A and A t draw, in the plane of the paper, two tan- 
gents to this sphere, intersecting at V. Consider V as the vertex 
of a cone tangent to 5; this cone will be a right cone cutting the 
plane of K in a certain conic K' with the same vertices A and 
A v Let the cone touch the sphere along the circle whose plane 
is T and which cuts the plane of K in a line perpendicular to the 
plane of the paper. This line appears as a point D. Assume 
any point P' on K! and let P'V cut T in Q; then P'Q = P'F (in 
space). The true length of P f F is P'R, which is parallel to VA. 
Now, no matter where P' is taken on K', P'R/P'D = P'F /P'D = 
constant. This constant is also equal AB/AD=AF/AD. 
Hence, as P' is the locus of the points whose distances from a 
fixed point F and a fixed line (D) have a constant ratio, it must 

1 Stated by Poncelet in the introduction of his Traite. It consists in the 
assumption that if one figure is obtained from another figure by a continuous 
variation, then projective properties derived from the first figure also hold for the 
second figure. The principle, however, is rigorous only when proved analytically. 



IS© 



PROJECTIVE GEOMETRY. 



be a conic with the focus F and the directrix (D) and is therefore 
identical with the original conic K. 

In Fig. 60 K has been assumed as an ellipse. For every 
sphere tangent at F there is consequently a right cone tangent 




Fig. 60. 
to it and of which the given ellipse is a section. Now 
VA l -VA = VB 1 +B l A 1 -VB-BA=B i A 1 -BA=A 1 F-AF = FF l ; 
i.e., VA t - VA = FF 1 = constant. 

V moves, therefore, on an hyperbola having A, A x as foci and 
F y F x as vertices. 



THEORY OF CONICS. 151 

If, in place of an ellipse, an hyperbola is chosen for K, V will 
be on an ellipse having the foci of the hyperbola as vertices and 
its vertices as foci. 

To sum up we have the theorem: 

The locus of the vertices of all right cones passing through 
a given ellipse is an hyperbola having the vertices of the ellipse 
as foci and its foci as vertices, and whose plane is perpendicular 
to the plane of the ellipse. 

The locus of the vertices of all right cones passing through a 
given hyperbola is an ellipse whose vertices and foci coincide with 
the foci and vertices of the ellipse, and whose planes are perpen- 
dicular to each other. 

Ex. Prove that the locus of the vertices of all right cones 
passing through a given parabola is a parabola having the vertex 
of the first as a focus and the focus as a vertex. The planes of 
the two parabolas are perpendicular. 

That there are no other right cones in these problems with 
the enumerated properties follows from the fact that in every 
right cone and one of its plane sections there is only one plane 
of symmetry with respect to the conic section. Conversely, if a 
conic is given, the vertex of a right cone can only be in this plane 
of symmetry, the plane passing through the foci and perpendicular 
to the plane of the conic. 

3. Perspective between any two given conies. 

Let K = o and K' = o be the equations of any two conies in 
the same plane. Apply a general perspective collineation to K, 
thus introducing five arbitrary parameters into the transformed 
equation. In order to make this last equation identical with 
K' = o, corresponding coefficients must be set equal. This gives 
five equations between the five parameters of the perspective col- 
lineation, and as these equations are of the second degree there 
will be several solutions of the problem. Two conies in a plane 
may therefore always be considered as perspectives of one another. 
Without discussing the possibilities of real and imaginary solu- 
tions of these equations the case will be considered where K 
and K f have four real points of intersection 1, 2, 3, 4, Fig. 61. 



152 



PROJECTIVE GEOMETRY. 



K and K' have the self -polar triangle XYZ in common. Desig- 
nate the four common tangents by I, II, III, IV, and consider 
the points of intersection 5 of III and IV, and S t of I and II. 
Evidently the centers of perspective must be sought in such 




Fig. 6i. 



points of intersection of common tangents, because a tangent 
from the center of perspective to one conic is also a tangent to 
the perspective conic. The common chords 12 and 34 as well 
as the chords of contact AB and A'W pass through X when pro- 
duced. Choosing 34 as the axis of a central collineation, S as 



THEORY OF CONICS. 153 

the center, A, A' as corresponding points, then the collineation 
is perfectly determined, and the conic K is transformed into a 
conic K" ', which, however, is identical with K', since it has the 
points 3, 4 and the points of tangency A', B' '; i.e., six points in 
common with K'. Instead of s we may also choose the chord s l 
as the axis, and S as the center of a collineation. Hence with 5 
as a center there are two central collineations transforming K 
into K' '. Conversely, every chord, as s, may serve for two collinea- 
tions with vS and S x as centers. The same can be analogously 
proved for every common chord and point of intersection of two 
common tangents. We have therefore the theorem : 

// two conies K and K f have jour real points 0) intersection, 
then there are 12 central collineations in which K and K f correspond 
to each other. For every point of intersection 0] two common 
tangents there are two chords which may be taken as axes 0] two 
0) those 12 collineations. Conversely, to every chord belong two 
points 0} intersection oj common tangents as centers 0) two such 
collineations. 

These propositions admit of an easy interpretation in space. 
As every common chord determines two centers of collineation, 
it follows that there are two cones through two conies in space with 
iwo points in common. 

In § 33 it has been shown that on account of the rectangular 
polar involution around the center of collineation not being 
changed, a circle concentric with the center of collineation is 
transformed into a conic whose focus is in this center. Gen- 
erally, for the same reason, a conic one of whose foci coincides 
with the center of collineation is transformed into a conic having 
the same focus. 

But this is also in agreement with the previous result. A 
focus of a conic may be considered as the point of intersection 
of two conjugate imaginary tangents from the circular points. 
Two conies with the same focus have therefore two common 
imaginary tangents, and their real point of intersection may 
be assumed as a center of collineation between the two conies. 

Ex. 1. Discuss the case and make the construction when 



*54 



PROJECTIVE GEOMETRY. 



K and K' intersect in two real points and have two parallel tan- 
gents. 

Ex. 2. Make the construction when iT.and K! are tangent. 

Ex. 3. Discuss the arrangements of K and K f in order to 
obtain all special cases of perspective collineation. 

4. Given five points of a conic K, — A, B, C, D, E; through 
two 0} these, say A and B, pass a circle K' } and find the center 0] 
perspective S for which K and K! are corresponding. 

In Fig. 62 construct the pole X of s or AB, which we assume 
as the axis of collineation. This is easily done by means of 
the points of intersection P and Q of CD and DE with 5 and 
their polars p and a in the quadrilaterals A BCD and AEBD. 
Construct also the pole X r of 5 with respect to K', then X and 
X r are corresponding points in the collineation and the center 5 




Fig. 62. 

must be on the line joining X with X '. DX and D'X' meet on 
s y in a point R; hence X'R cuts K' in D' . Joining DD' and 
producing gives on XX' the required center 5. Having found S, 
it is an easy matter to construct further elements of K from the 
corresponding elements of K! . 



THEORY OF CONICS. 



155 



A similar problem was solved in the first part of this section. 

5. Given three points A, B, C of a conic K and the tangents 
at A and B. To find the point of intersection of this conic with 
a given line g. 

Designate the intersection of the tangents by S,. Fig. 63, and 
draw any circle K' tangent to SA and SB at the points A' and 
B' . K and K' are now corresponding conies in a collineation 
with S as a center and A , A ' ; B } B' as corresponding pairs. SC 




Fig. 63. 



cuts K' in C. Draw CA and CB cutting g in X and Y. Draw 
SX and SY, which by C'A' and CB' are cut in X' and Y' . The 
line joining X'Y' is g'. Let g' cut 2T' in P r and Q' ', then SP' 
and SQ' produced cut g in P and <2, the points of intersection 
of g with K. 

With exactly the same designations we can immediately 
solve the special case: 

Given the asymptotes of an hyperbola and another point. To 
find the points of intersection of this hyperbola with a given straight 
line. 



156 



PROJECTIVE GEOMETRY. 



Nothing is changed in the previous construction except that 
A and B are the infinite points on the asymptotes. 

6. To construct a conic K when three points A, B, C and two 
tangents a and b are given. 

Draw again a circle K f tangent to a and b and assume the 
intersection S of a and b as the center of collineation, Fig. 64. 
Join S to A, B, C and designate the points of intersection of 
SA, SB, SC with K' by A', B' , C and A", B" , C". If we let 




Zz?r 



Fig. 64. 

A' j B' , C r correspond to A, B, C, then 5 is the axis of collineation; 
if the corresponding points are A", B", C" , then s x will be the 
axis of collineation. In both collineations the same conic K 
corresponds to K' '. But we may also let A', B", C correspond 
to A y By C, which will lead to a different conic A". The arrange- 
ment A"B'C n ', ABC leads to the same conic. There are 
eight different correspondences possible which in groups of 
two lead to the same conic. The problem admits, therefore, of 
jour different solutions. 



THEORY OF COXICS. 



157 



7. Given three points of a conic K, — A, B, C, — and a focus 5. 
To construct the conic. 

Draw in Fig. 65 any circle tangent to the conjugate imaginary 
tangents from 5 to K\ i.e., draw anv circle K! with 5 as a center. 




The problem and the construction are in this case exactly the 
same as in problem 6. Here also there are four different solutions , 

8. Given jour points and a tangent of a conic, to construct it. 

In Fig. 66 let A , B, C, D be the given points and t the given 
tangent of the conic K. Consider i5 or ^ as the axis of a col- 
lineation, and any circle K' through A and B as the perspective 
of K. By means of the quadrilateral A BCD construct the polar 
p of P with respect to K, and also the polar p' of P with respect 
to K f . p and p' meet in a point of s. From the point where t 
cuts 5 draw the tangent t f to K' and let this tangent correspond 
to t in the collineation. t and t ' cut p and p' in two correspond- 
ing points X and X' , and the center or, if there are several, the 
centers of possible collineations must be on the line joining X 
with X' . To determine these centers, join C with X, and the 
point of intersection of CX with 5 to X'. Where the last line cuts 



*S* 



PROJECTIVE GEOMETRY. 



K r are two points C and C/ which correspond to C. Hence there 
are two different collineations. Their centers 5 and S 1 are ob- 







Fig. 66. 



tained as intersections of C'C and C/C with XX'. To the point of 
tangency V oi t f with K' correspond in the two collineations T 



THEORY OF CONICS. 159 

and 7\ on t, which are the points of tangency of the two conies K 
passing through A BCD and tangent to t: 

That these are the only solutions is not apparent from this 
construction; it simply shows how conies with the required con- 
ditions may be found. 

9. Osculating Circle 0] a Conic. 

If a conic K passes through the center S of the collineation, 
then K! will be tangent to K at S. If, furthermore, also the axis 
5 passes through S, one of the remaining two points of intersection 
of K and K f will coincide with S, and K and K r have at 5 a con- 
tact of the second order. If K is a circle, K will be the osculat- 
ing circle to K f at S. The remaining fourth point of intersection 
will be on s. In case that 5 is on s, the counter-axes of collinea- 
tion a' and r will be on opposite sides of s. The center M of the 
circle K is the pole of the infinitely distant line q with respect to 
K. The corresponding point M f of the collineation is the pole of 
q f with respect to K' '. If now a diameter of K turns, about M, 
the rays joining S with its extremities form a rectangular involu- 
tion of rays around 5 which is identical with the involution of rays 
joining S to the extremities of the chords through M' correspond- 
ing to the diameters through M. Hence M' is obtained as the 
pole of the involution of points on K', which when joined with S 
give a rectangular involution of rays. q f is the polar of M' with 
respect to K', and 5 is a line through 5 parallel to q f . 

It is now possible to solve the problem : Given -five points 0) a 
conic K f , to construct the osculating circle at any of the given 
points. 

In Fig. 67 let A, B, C, D, E be the given points, and A 
the point . at which the osculating circle shall be constructed. 
Join A with B and C; at A erect perpendiculars to AB and AC, 
and by Pascal's theorem construct the intersections B 1 and C\ 
of these perpendiculars with K' '. BB X and CC t cut each other 
at M' '. In the quadrilateral BB£C X the polar q' of M' is easily 
found. Through A draw s parallel to q', and find by Pascal's 
theorem the intersection F of 5 with K' ' . K is the circle passing 
through F and tangent at A to the perpendicular to AM'. 



i6o 



PROJECTIVE GEOMETRY. 



If K is tangent to s at the point S, the osculation will be of 
the third order; i.e., K and K r will have four points in common 
at 5. 

Ex. i. To construct a conic when the osculating circle at 
one of its points, A, and two other points, B and C, are given. 




Fig. 67. 



Ex. 2. To construct the osculating circle at the vertex of a 
conic which is determined by major and minor axes. 

Ex. 3. To construct a parabola when the osculating circle at 
its vertex is known. 

Ex. 4. Given three tangents and two points of a conic; con- 
struct the conic. Dualistic problem of § 42, 6. 



Ex. 5. Given four tangents and a point of a conic: 
the conic. Dualistic problem of § 42, 8. 



construct 



THEORY OF CONICS. 161 



§ 43. Problems of the Second Order. 

1. In the previous section and even farther back we have 
occasionally touched upon problems of the second degree. We 
shall now pay particular attention to a few geometrical problems 
which analytically are equivalent with the solution of an equation 
of the second degree. Most of these problems may be reduced 
to the problem, to find the double-points of two coincident projective 
point-ranges. This was done analytically in the first chapter. 
Geometrically we may solve it by the following proposition, which 
in a little different form appears as Pascal's theorem: Six points, 
A y B, C, A', B f , C, on a conic K determine two projective ranges 
of points on K, so that for any point P on K we have the pro- 
jective pencils 

P(ABC . . .)=P(A'B'C ...). 

The pairs of sides AB f , A'B; BC, B'C\ CA f , CA intersect in 
three collinear points, on the Pascal line p. Considering two 
points, for instance B and B', as carriers of pencils, then 

(B-A'B'C .. .) = (B'-ABC . . .), 

and as BB f is a ray common to both pencils, they are perspective 
and have p as the axis of perspective. Two rays joining B and 
B r with any point on p cut K in two corresponding points of the 
projective ranges on K. From this it is clear that the points of 
intersection of p with K are the double-points of the projectivity. 
These are real, coincident, or imaginary according as p cuts, 
touches, or does not cut K. 

If, instead of six points, six tangents a, b, c, a f , b f , d of the 
conic are given, the lines joining the pairs of intersection ab', 
a f b\ be', b'c\ caf, c'a all pass through the Brianchon point P. 
Considering two of the tangents, b and b f , and cutting these with 
the remaining tangents, two perspective ranges 

(b-a'b'c' ...) = (b'-abc.) 



162 



PROJECTIVE GEOMETRY. 



with the common point W (self-corresponding) are obtained. 
Taking any ray through P and cutting it by b and b', then the 
tangents to K from these two points are two corresponding tan- 
gents. The tangents jrom P to K are the double tangents 0} the 
projectivity. We postulate now that a problem is geometrically 
solvable if it can be solved by compass and ruler. Hence, replac- 
ing K by a circle, we can now solve the following problems: 




Fig. 68. 



2. Given five points oj a conic; to construct the intersections 
0} this conic with a straight line. 

In Fig. 68 let AJBJOJ)^ be the given points and I the 
given line. Join D t and E x with A 1 B 1 C 1 and designate the 



THEORY OF CONICS. 



163 



corresponding intersections on / by ABC and A'WC . Draw 
an arbitrary circle K and join any of its points S with ABC and 
A'B'C ', and let these lines cut K in points which we shall also 
designate by ABC and A'B'C, to simplify the designation. 
According to the foregoing results ABC and A'B'C determine 
two projective ranges on K. Construct the line p and let M 
and N be the intersections of p with K. Join 5 with M and N 
on K, and produce to their like-named intersections M and 
N on L These will be the double-points of the projective 




Fig. 69. 

ranges ABC . . . and A'B'C , or the points of intersection of / 
with the conic through ABCDE. 

3. Given five tangents of a conic; to construct the tangents 
0] this conic through a given point. 



1 64 PROJECTIVE GEOMETRY. 

In Fig. 69 let a 19 b ly c^ d v e 1 be the given tangents and S the 
given point. Cut d x and e x with aj}^ and designate the corre- 
sponding hnes, joining these points with S, by a b c 2 a' b' c' '. Draw 
an arbitrary circle K and let any tangent s oi K cut abca'b'c' in 
six points. From these draw tangents to K and designate them 
similarly by abc, a'b'd '. Construct the Brianchon point P of the 
circumscribed hexagon abca'b'c' of K. From P draw the tan- 
gents m and n to i£, cutting 5 in M and iV. The lines joining 6* 
with M and N are the required tangents from S to the given 
conic. 

This construction may be replaced by a simpler one. Join 
the points where a ly b lt c x cut d x and e x directly to the point S of 
an auxiliary circle passing through S, and designate the points 
of intersection with this circle by A, B, C, A', B', C. Con- 
struct the Pascal line p of this hexagon. The lines joining S with 
the points of intersection of p with the auxiliary circle are the 
required tangents from S. 

4. Poncelet's Problem. — To construct a polygon whose 
vertices shall lie on given straight lines (each on each), and whose 
sides shall pass through given points (each through each). 

For the sake of simplicity we shall limit the problem to four 
straight lines a, b, c, d and four points A, B, C, D. The method 
of reasoning is not different in the general case. First make a 
trial construction by drawing through A a line a x cutting a in A ly 
Fig. 70. From A x draw a line b x through B, cutting b in J5 X ; 
from B x a line c x through C, cutting c in C x \ from C, a line d x 
through D, cutting d in D t ; from D x a line a/ through A cutting 
a in A{. If a x turns about A, then b v c lt d lt a{ will turn about 
B, C, D, A in such a manner that we have for various positions 
the projective ranges 

(A x A 2 Az • • :) = (B t B 2 B 3 . . .) = (C 1 C 2 C 3 . . •) = 

(ADA-..) = (i/i/4/...). 

Considering the projective ranges 

(A X A 2 A 3 . . .) = (A j A 2 A 3 . . .) 



THEORY OF CONICS. 



165 



on a, it is clear that the lines a m a N through A and the double- 
points of these ranges, M and N, coincide with the lines a M ', 
a N ' through D. These double-points determine, therefore, two 
solutions of the problem which may be real or imaginary. In 




the figure the two real solutions of the problem are indicated by 
heavy lines. 

Ex. 1. To inscribe in a given conic a polygon whose sides 
pass respectively through given, non-collinear, points. 



1 66 PROJECTIVE GEOMETRY. 

Ex. 2. To circumscribe about a given circle a triangle whose 
vertices are on three given lines. 

Ex. 3. Between two given straight lines to place a segment 
such that it shall subtend given angles at two given points. 

Ex. 4. To construct a polygon whose sides shall pass respect- 
ively through given points, and all whose vertices except one 
shall lie respectively on given straight lines; and which shall be 
such that the angle included by the sides which meet in the last 
vertex is equal to a given angle. (Cremona.) 

Let A, B, C, . . . , N be the given points and a, b, c, . . . , m 
the given lines. Through A and N draw a circle K which sub- 
tends the given angle over the chord AN. From any point of K 
draw a line through A, cutting a in A t ; from A x draw a line 
through B, cutting b in B 1} and so forth, until the line m is reached 
in a point M x . Then through N and the. same point on K draw 
a line cutting m in M-[. Repeat this construction for two other 
points of K, thus giving on m the projective ranges (A 1 A 2 A 3 . . .) = 
{A{A 2 'A 3 . . .). The double-points of these ranges make it 
possible to draw two polygons with the required properties. 
This problem may be solved in a different manner. 

Through A draw any line a x cutting a in A ly through A x and 
B a line b 1} and so forth, until the line m is reached in M x . 
Through M x and N draw a line n 1 cutting a x in a point V. If 
now a x turns about A, then n x will turn projectively about N. 
Hence their point of intersection V will describe a conic K* 
passing through A and N. The two other points of intersection 
of this conic with the circle K determine the two solutions of 
the problem. It is now possible that the conic K* is itself a 
circle, but different from K. In this case there is no real solu- 
tion. K* may be identical with K, so that there are an infinite 
number of solutions. 

Make the constructions as indicated. 



THEORY -OF CONICS. 



167 



§ 44. An Optical Problem. 

1. The following problem stated by Cremona in his Elements 
of Projective Geometry, p. 199, is an application of Ex. 4 of the 
previous section: 

A ray of light emanating from a given point A is reflected 
from n given straight lines in succession; to determine the original 
direction which the ray must have, in order that this may make 
with its direction after the last reflection a given angle. 




Designate in Fig. 74 the reflecting lines by a, b, c, . . . , n. 
Through A draw any ray a 1 striking a at A v The reflected 



1 68 PROJECTIVE GEOMETRY. 

ray b x passes through the point B which is symmetrical to A 
with respect to a. The ray b t strikes b at B ly and its reflected 
ray c x passes through C, which is symmetrical to B with respect 
to b, and so forth. The ray o 1 reflected from the last line n 
at N 1 passes through O, which is symmetrical to N with respect 
to n. Let the rays a ± and o x intersect at V v We have now a 
closed polygon a 1 b 1 c 1 . . . o 19 whose sides pass through the fixed 
points A , B, C, . . . N, O and whose vertices except V lie on the 
fixed sides a,b, c, . . . , n. Hence, when a x turns about A , V will 
describe a conic and the problem is reduced to the one explained 
in Ex. 4, § 43. 

2. Cremona stops the discussion of the problem at this point. 
We shall now show that a further investigation is necessary. 
Let a be the angle of incidence of a^ on a, a 1 the angle of inci- 
dence of b 1 on b, a 2 of c 1 on c, and so forth; <j> ly cj) 2 , cf) 3 . . . the angles 
which a and b, b and c, c and d, . . . include. The angles <f> 1} <j> 2 , 
</> 3 , . . . between the different reflecting lines must be selected in 
such a manner that always 

a,i-\-a i+1 +(f)i = 7r. 
From the figure we now derive the following series: 



«1= 


a, 








a 2 = 7i- 


-a-fa, 








«3 = 


a+<£i-<£ 2 , 








a 4 = 7i- 


-a-<j) 1 +(/> 2 - 


"03. 






«5 = 


a+&— 0a-+03- 


"04> 




a 6 = 7r- 


-<*-0i+02- 


-03+04" 


'05> 



a 2il = 7Z—a— 0!+0 2 — 03+ • • • — 02j«-l> 

a 2 ^ +1 = a + 0i— 02+03— • • • — 02/*- 

If there are w reflecting lines, then the number of angles a 
is also n, and the number of angles is w— 1. Erect perpendicu- 



THEORY OF CONICS. 169 

lars to a ± at any of its points, to a at A v to b at B lt and so forth, 
to n at any of its points. Then the first and the last perpen- 
dicular deviate from each other by the angle 

a+(n-2)7i+a n — (0i+0 2 +0 3 + • • • +<£«-i)> 

and consequently also the rays a x and o x by the angle 0; 

0=a + an-(0i+0 2 +0 3 + ... + n _ 1 )+(«— 2>r. 

If n is odd, then n— 1 is even; i.e., n = 2/1+1, and 

=a + a:+ (j> 1 — 2 + 3 — ...— 0^— X — 2 — 3 — ...— 2/ <+ (n— 2)n, 

or 

= (n- 2 )n+ 2a- 2 (0 2 + 2 + 4 + ... + 2/t ) ; 

i.e., the angle between the original incident ray of light and the 
final emanant ray depends upon the angles and the original 
angle of incidence a. Their point of intersection V describes a 
conic which is not a circle, and there are two positions of V, real 
or imaginary, for which the incident and reflected ray make a 
given angle. 

If n is even, then n— 1 is odd; i.e., 71 = 2/1 and 

0=«+7T -« -0 1 + 2 -0 3 +... -0^_i-0i-02-. ..-0 2 /i-l+(W-2)^, 

or 

0=(rc-l)7r-2(0 1 +03+... + 2/t _ 1 ). 

Hence, in case oj an even number of reflecting sides, V describes 
a circle and the angle is constant. Cremona's problem admits 
either oj no solution, or oj an infinite number oj solutions. The 
angle does not depend upon the angles oj even indices. 

To sum up we may state Cremona's problem and its solution 
by the following proposition: 

// rays oj light emanate jrom a fixed source which, in succession, 
are reflected on n straight lines, then the last reflected rays cut the 



170 



PROJECTIVE GEOMETRY. 



corresponding original rays in points oj a conic, which is not a 
circle, when n is odd, and in points oj a circle when n is even. In 




Fig. 72. 

the first case there are two places on the conic at which the original 
and the final ray make a given angle. In the second case there are 
no such places on the circle, or else an infinite number. In this 



)<f>=90 




Fig. 73. 



case the angle <f> depends only upon the angles between succeeding 
reflecting lines whose orders in this succession are odd. 
3. Applications. — Let n = 2, then <£ = 7r— 2^. 



To make <£=— , we must choose 

2 



45 . This case is illus- 



THEORY OF CONICS. 171 

trated in Fig. 72 and is practically applied in Bauernfeind's Angle 
Mirror or Optical Square. 
For n = 4 

^ = 371-2(^+^3). 

To make <f> =—, <£i+<£ 3 must be made equal to — n. Under 
this condition <f> x and <f> 3 may vary separately. The condition 
is also satisfied by taking (j) 1 = <j> 2 =-^7t=ii2° 30', and this is 
illustrated in Fig. 73. 



CHAPTER IV. 

PENCILS AND RANGES OF CONICS. THE STEINERIAN TRANS- 
FORMATION. CUBICS. 

§ 45. Pencils and Ranges of Conies. 
1. Involution of the Pencil u+hi 1 (1) = o. 

Let u=ax 2 -\-2bxy-\-cy 2 +2dx+2eyi-f = o, 

u x = a^ 2 + 2b 1 xy+ c x y 2 -\- 2& x x-\- 2e 1 y+ j x = o 

be the equations of two conies; then 
(1) u-\-Xu 1 = o 

is the equation of a conic which passes through the four points 
of intersection of U and U v As a conic is determined by five 
points, any fifth point, different from one of the four points of 
intersection of U and U ly determines the equation of the conic 
through the five points; i.e., A. Conversely, every value of X 
determines the equation of one of the conies of the system. 
Designating the points of intersection of U and U l by A, B, C, D, 
then for a variable A, 

u-\-Xu 1 = o 

represents all conies through the four fixed points A, B, C, D, 
and is called the equation of the pencil 0} conies through these 
points. Among these conies are three degenerate conies, con- 

1 See Joachimsthal, loc. cit., p. 183. By U, V, U v etc., we shall designate 
conies whose equations are u=o, v=o, u t = o, etc. 

The student is asked to draw a figure for this section. 

172 



PENCILS AND RANGES OF CONICS. 173 

isting of the three pairs of lines through A, B, C, D. To prove 
this we form the discriminant of (1), which is 



(2) 



a+Xcii b+Xb x d-\-Xd 1 
b-{-Xb 1 c J r Xc 1 e-\-Xe 1 
d+Xd ± e+Xe 1 }+Xf 1 



The vanishing of this expression is the condition for degener- 
ate conies among the pencils. This gives a cubic equation in X 
and consequently three values for X; i.e., three degenerate conies 
through A, B, C, D, as was to be proved. One value of X is 
always real, so that also in case of one or two imaginary pairs 
among A, B, C, D there is always a conic consisting of a real 
line-pair. In case of a double- root which is evidently real, the 
third root is also real; the conies U and U x have a contact of 
the first order. If the root is triple, U and U 1 have a contact of 
the second or third order. 

In § 36 it was shown that the coordinates of a point C (x, y) 
on the line joining the two points A(x lf y t ), B(x 2 , y 2 ) are 

x x -Xx 2 yi~b'2 1 .AC 

* = ^— T' y = ^ T , where X = m 

Assume now that A and B are on the conic given by (1), 

AC 

then C is on the conic U, if X = -^ is a root of 

(3) u i —2Xv+X 2 u 2 = o, 

where u 19 v, u 2 have the same meaning as in formula (2), § 36. 
In a similar manner, C is on Z7 1? if X satisfies 

(4) u 1 '—2Xv' + X 2 u 2 =o, 

where w/, 1/, u 2 have the same meaning as in (3), except that 
a, b, c, . . . are replaced by a lt b lt e lt . . . . Designating the points 



174 PROJECTIVE GEOMETRY. 

of intersections of AB produced with U by C, C'\ and with U l 
by C lt C/, we have 

AC AC__u 1 AC, AC£_u/ 
BCBC'~uj BC x 'BCl~uj' 

The points A and B are on u-\-Xu x = o\ hence 

and, eliminating X, 

u x u/ 

— = —,, or 

(5 ^ BC'BC'~ BCiBCf 

Giving A all possible values and keeping the transversal, 
or C, C f and C 1? C/, fixed, yl, J3 is the pair of points in which 
the variable conic u-\-Xu 1 = o cuts this transversal. (5) may 
also be written 

(ABCC 1 ) = (BAC'C 1 ')) 

i.e., the anharmonic ratio of any four points cut out by the fixed 
and variable conies on the transversal is equal to the anharmonic 
ratio of the four corresponding points. Furthermore, from (5) 
it is seen that interchanging A and B, two corresponding points, 
does not affect the relation. The system of points defined by (5) 
is therefore involutoric. Hence the theorem: 

The conies oj a pencil oj conies cut any transversal in an involu- 
tion oj points. Every conic, including the degenerate conies, cuts 
out a pair of the involution. (Desargues.) 

The double- points of the involution are evidently the points 
where two conies of the pencil touch the transversal. They may 
be real (including coincidence) or imaginary. The remark in 



PENCILS AND RANGES OF CONICS. 175 

connection with problem 8, § 42, to construct a conic through 
four given points, tangent to a given line, is now clear. 

Corollary. — Any transversal cuts the three pairs of sides 
of a complete quadrilateral in three pairs of an involution. 

Ex. Prove directly that every transversal cuts a coaxial system 
of circles in an involution. By reciprocation we derive the 
theorem : 

The pairs of tangents from any point to the conies of a range 
iconics inscribed in a quadrilateral) form an involutoric pencil. 

Corollary. — The lines joining any point with the three 
pairs of vertices of a complete quadrilateral form three pairs of 
an involutoric pencil. 

Ex. Prove directly that the tangents from any point to the 
range of circles inscribed to two straight lines form an involutoric 
pencil. 

2. A Special Case. — Assume the four points A, B, C, D, 
through which the pencil of conies passes, as an orthogonal quadri- 
lateral; i.e., AB±CD, BC LAD, CA ±BD. In this case the 
conies are all equilateral hyperbolas. To prove this, note that the 
degenerate conies consist of three pairs of perpendicular lines, 
Fig. 74. The involution on the infinitely distant line is there- 
fore rectangular and its pairs can only be cut out by conies 
whose infinite branches are rectangular; i.e., branches of equi- 
lateral hyperbolas. 

Take now any equilateral hyperbola and on it any triangle 
ABC. Let D be the point of concurrence of the altitudes of 
ABC. Through A BCD we can now pass an infinite number 
of equilateral hyperbolas, among which is necessarily the given 
hyperbola. Hence D is on this hyperbola, and we have the 
theorem : 

The point of concurrence of the altitudes of any triangle in- 
scribed in an equilateral hyperbola lies on this hyperbola. 1 

3. Polar s of a Pencil of Conies. 

1 Brianchon et Poncelet in Gergonne's Annales, Vol. II, pp. 205-220. Also 
Fiedler in Vierteljahrsschrift d. Naturf.-Ges., Zurich, Vol. XXX, pp. 390-402. 



176 



PROJECTIVE GEOMETRY. 



From the explicit expression of the equation of a pencil of 
conies 



(6) 



u-\-Xu 1 = o 



it is easily found that the equation of the polar of a point P may 
always be put in the form 



(7) 



p+Xp^o, 



where p = o, p 1 = o are the equations of the polars of P with 
respect to the conies U and U v From this it follows that the 




Fig. 74. 

polar of any point P with respect to a conic of the pencil always 
passes through the point of intersection P r of the polars of P 
with respect to U and U v Hence the theorem: 

All polars of a point with respect to the conies 0] a pencil are 
concurrent. 

If the point P with the coordinates x ly y x describes the straight 



PENCILS AND RANGES OF CONICS. 



177 



line a#i+/fyi+7- = o, then for the point P f = (x 1 , y 1 f ) we nave 
the three conditions 

(ax/ -}- by I + d)x x + (bx/ + cy{ + e)y 1 -\- dx/ + ey{ + / =0, 

(a^/ + b x y{ + dj^ + (6i x{ + c^/ + e/)^ + d x x{ + ^v/ + / x = o,, 

which are consistent only when 



(8) 



ax/ +63// +d bx ± -\-cyl +e dx( +ey 1 ' + f. 

a % %l + b x y{ + d x b x x{ + c^/ + e 1 d x x{ + ^v/ + f 1 

a ft r 



This gives a quadratic equation between x/, y/, the coor- 
dinates of P' ; hence the theorem: 

// a point P describes a straight line, then the point of con- 
currence P r of all polars with respect to the conies of a pencil 
describes a conic. 

Designating ax^+by^ + d, bx^+cy^+e, dx/ -{- ey/ -\- f by r , s, t, 
and in a similar manner by r 1} s 19 t x the same expressions, v/ith 
a, b, e, . . . replaced by a lt b ly c lf the polar of a point (x t f , y±) with 
respect to the pencil of conies u-\-Xu 1 = o has the form 

(r+XrJx^ (s+Xs 1 )y 1 + t+Xt 1 =o. 

This equation will be identical with that of the given line g, 

ax 1 +Py 1 +Y = o, if 

r+\r x s+Xs t t+Xt t 



(9) 



a 



P 



Hence the pole (x/, v/) of g for any conic of the pencil must 
satisfy (9). For every value of X a definite value of (x/, v/) 
is obtained, which therefore describes a certain locus. To find 



178 PROJECTIVE GEOMETRY. 

its equation we must eliminate X from (9), which gives the equa- 
tion 

(10) a(^ 1 -5 1 /)+/3(r 1 /-r/ 1 ) + r(^i-r 1 5) = o. 

But this is identical with (8). Hence the theorem: 

The locus of the poles of a straight line with respect to all 

conies of a pencil is a conic which is identical with the conic of 

concurrent polars of all points of the given line with respect to 

the same pencil. 

Of particular interest is the case when P describes the infi- 
ll 

nitely distant line; i.e., when — • = /* is a variable (finite) and 

x t = 00. In this case 

(p= ax( + by ± ' + d + p.(bxj + cy{ + e ) = 0, 

•^ ( pt= <*!#/+ &1V+ d i+ K b i x i+ c iyi+ e i)=o. 

Eliminating /z, the equation of the conic which P r describes 
becomes 

(1 2) {ax/ + by J + d) (b.x/ + c${ + e t ) 

- (a^/ + b x y x f + d t ) (bx/ + cy{ + e) = o. 

If (#!, v x ) describes the infinitely distant line, then its pole 
with respect to a certain conic of u-[-Xu 1 = o must satisfy 
p+Xp 1 = o for all values of p., which can only be true when equa- 
tion (12) is satisfied. But the poles of the infinite line are the 
middle points of the conies of the pencil. The centers of a pencil 
of conies lie on a conic whose equation is given by (12). 

The three diagonal points of the fundamental quadrilateral, 
being the centers of the three degenerate conies, belong to this 
locus. If P is taken as the infinitely distant point of the line 
joining two points, say A and B, of the fundamental quadri- 
lateral, then P' is the middle point of AB, since all polars of P 
with respect to the conies of the pencil pass through this point. 
The locus (12) passes, therefore, also through the middle points 



PENCILS AND RANGES OF CONICS. 179 

of AB, BC, CD, DA, BD, CA. In case of an orthogonal quad- 
ruple, as it was described above, under (2), the locus (12) becomes 
a circle circumscribed to the foot-points of the altitudes of the 
triangle A BC, which bisects the sides and the segments DA, DB, 
DC of the altitudes, Fig. 74. This circle is otherwise called 
the Feuerbach circle 1 of the triangle. We have therefore the 
theorem : 

The locus of the centers of all equilateral hyperbolas circum- 
scribed to an orthogonal quadrilateral A BCD (D= point of concur- 
rence 0) altitudes) is the Feuerbach circle 0] the triangle ABC. 

4- Poles 0} a Range 0} Conies. 

A range of conies consists of all conies inscribed to a quadri- 
lateral (imaginary elements included), or is the reciprocal of 
a pencil of conies w+Aw i = o with respect to a given conic K. 
From this property we derive immediately the theorems : 

A II poles of a straight line with respect to the comes of a range 
are collinear. 

If a straight line p turns about a fixed point, then the line of 
collinearity p f of all its poles with respect to the conies of a range 
envelops a conic. 

Let M be the center of K and designate by V a conic of 
the pencil u-\-Xu l = o J and by v the polar of M with respect 
to V. On reciprocation with respect to K, V is transformed 
into a conic V ; M, the pole of v, is transformed to infinity; 
consequently the polar v is transformed into the center of the 
transformed conic V '. As the polars of M with respect to 
all conies V are concurrent, it follows that their reciprocal poles are 
collinear. 

Hence the theorem: 

The centers of the conies of a range are collinear. 

Among the conies of the range there are three degenerate 
ones, consisting of the three pairs of points in which the sides 
of the fundamental quadrilateral intersect each other. The 

1 Concerning this circle see Cajori's History of Elementary Mathematits, pp- 
259, 260; also Kotter: Die Entwickelung der synthetischen Geometric, Vol. I, 
PP- 35~3 8 - 



180 PROJECTIVE GEOMETRY. 

middle points of these three pairs evidently belong to the above 
locus. This may be stated in the corollary: 

The middle points of the three diagonals o) a complete quadri- 
lateral are collinear. 

Designating the three pairs of points by A^A 2 , B 1 B 2 , C X C 2 , 
two circles over A X A 2 and B 1 B 2 as diameters intersect at two 
points (real or imaginary) F ly F 2 . Joining F 1 toA li A 2 and B lt B 2 , 
the involution of the tangents from F 1 to all conies of the range 
is determined, and as two of these pairs, F X A X , F X A 2 and F X B 1: 
F X B 2 , are rectangular, all other pairs are rectangular; i.e., 
F 1 C 1 A.Ffi 2 , and the circle over C X C 2 as a diameter is coaxial 
with the first two circles. 1 Constructing all circles from the 
points of which rectangular pairs of tangents may be drawn to 
the conies, it follows from the last remarks that all these circles 
form a coaxial system. We state these facts once more in the 
theorem : 

The circles from whose points pairs of perpendicular tangents 
may be drawn to the conies oj a range, each jor each, form a coaxial 
system. 

Ex. i. If through the vertices of two angles, whose sides 
intersect each other in the points A, B, C, D, two parallel lines 
are drawn, then the harmonic lines of each of these parallel lines 
with respect to the sides of the corresponding angles intersect 
each other in a point which describes the conic of the middle 
points of all conies through A, B, C, D, when the direction of 
the two parallel lines changes. 

§ 46. Products of Pencils and Ranges of Conies. 

1. The pencils and ranges of conies may be related to each 
other by requiring that two conies shall correspond to each other 
if their equations are determined by one and the same param- 
eter A. Thus, for a certain value of X, 

u+Xu x =0, 



(l) 1 v+Xv t =0 



1 The student is asked to make the foregoing construction. 



PENCILS AND RANGES OF CONICS. 181 

represent two corresponding conies. The two pencils (i) are 
said to be projective. Two corresponding conies intersect each 
other in four points (including imaginary points). We obtain 
the locus of all these points by eliminating A from equations (i), 
which gives 

(2) uv 1 —u l v = o, 

an equation of the fourth degree in % and y. Hence the theorem : 

Tivo projective pencils of conies produce a curve of the fourth 
order. 

As the equation of a curve of the fourth order depends upon 
twelve constants and as (2) contains twenty constants, it is 
evidently always possible to state the converse; i.e., 

Every curue 0) the fourth order may be considered as the product 
oj two projective pencils of conies. 

The curve as represented by (2) passes through the inter- 
sections of U and U lt V and V lt U and F, U 1 and V v 

Reciprocally we have the theorems: 

Tivo projective ranges of conies produce a curve of the fourth 
class. 

Every curve of the fourth class may be considered as the product 
of two projective ranges of conies. 

2. In analogy to (1) a pencil of conies and a pencil of rays 
are projective if their equations may be written in the respective 
forms 

( u+ki t =0, 

{3) ii+* =0. 

For every / we have a conic and a ray corresponding to each 
other in this projectivity, and the two intersect each other in 
two points. The locus of these points is obtained by eliminating 
A from equations (3); so that its equation is 

(4) ul^-u x l = o, 



1 82 PROJECTIVE GEOMETRY. 

and is of the third degree. It is satisfied for l = o, l l = o; u=o, 
^=0; u = o, / = o; % = o, l t = o. Hence the theorem: 

The product of a pencil of conies and a projective pencil of 
rays is a curve of the third order which passes through the vertex 
of the pencil of rays and through the four fundamental points of 
the pencil of conies. 

As the equation of a cubic depends upon nine constants and 
as (4) contains fourteen constants, it is always possible to write 
the equation of any cubic in the form of (4). Hence the theorem: 

Every cubic may be considered as the product of a pencil of 
conies and a projective pencil of rays. 

Reciprocally : 

The product of a range of conies and a range of points is a 
curve of the third class which is inscribed to the fundamental 
quadrilateral of the range of conies and which touches the range 
of points. 

Conversely, every curve of the third class may be considered 
as such a product. 

3. In § 45, 3, it was shown that the polars of a point P with 
respect to a pencil of conies u+Xu^o are concurrent at a 
point P', and that when P describes a straight line, P' describes 
a conic. In general to a point P corresponds one and only one 
point P' . Let the straight line described by P be g and the 
corresponding conic described by P f be G, and designate the 
points where g cuts G by X and X', Fig. 75. The relation between 
P and P r is involutoric; i.e., all polars of P' with respect to the 
pencil pass through P. 

To the point X on g corresponds a point on G, to X on G corre- 
sponds a point on g ; but to X only one point corresponds in the 
correspondence between P and P' '; hence the point correspond- 
ing to X is X'. Conversely, to X' corresponds X. The pencil of 
conies cuts g in an involution of points. Let M and N be the 
double-points of this involution, V M and V N the conies of the 
pencil touching g at M and N. Then, the polars of M with 
respect to V M and V N are g and the polar passing through N. 
Hence, in the correspondence of P and P', to M corresponds 



PENCILS AND RANGES OF CONICS. 1&3 

me point N, and, conversely, to N corresponds M. There are 
only two points on g with this property, the points X and X' where 
g cuts the conic G. Consequently M and N coincide with X 
and X' '. We may state this result in the theorem: 



Fig. 75. 

In the correspondence of P and P' to a straight line g corre- 
sponds a conic G. The points on g whose corresponding points 
are on g itself are the points of intersection X and X' of g with G. 
These same points are also the double-points of the involution of 
points which the pencil of conies cuts out on g. 

According to the theorem that G is also the locus of the poles 
of g with respect to the conies of the pencil, the points X and X f 
on G are poles of g, and as these coincide with g it follows that 
g touches two conies of the pencil at X and X f ,\ in other words, 
X and X' are the double-points of the involution cut out on g 
by the pencil of conies, as has been established above. The 
theorem therefore also holds for an imaginary pair of corre- 
sponding points X, X'. 

4. Consider now the straight lines of a pencil: 

(5) (<*+ fia 1 )x+ (p+ upOyi- r + Wi=o. 



1 84 



PROJECTIVE GEOMETRY. 



For a definite value p. we have a definite ray of the pencil. 
According to (8) in § 45, 3, when P describes the .line (5), P' 
describes the conic 



(6) 



ax/ + by/ + d 
a 1 x/+b 1 y/+d 1 



bx/ + cy/ + e 
K^i J rC 1 y/-\-e 1 



dx/+ey/ + j 
d 1 x/+e 1 y 1 '+f 1 



o, 



which may also be written in the form 

(7) 



ax/ + by/ + d bx/ + cy/ + e dx/ + ey/ -f / 
a x x/ + Z^v/ + d x b x x/ + qv/ + e 1 d x x/ + ^v/ + j x 
a P • "r 



+ V 



ax/ + Z>v/ + d &#/ + cy/ + e dx/ -f ev/ + / 
a x x/ + b x y/ + ^ fr^/ + c x y/ + ^ d^/ + e x y/ + / t 



o. 



Designating ax+/?v+ 7- and a^+ft^H- ft by g and & and the 
corresponding conies by G = o and G x = o, then to the pencil 
g+ f J gi = ° corresponds the projective pencil of conies G+//G 1 = o. 
The product of the two pencils is therefore a curve of the third 
order with the equation 



(8) 



Gg 1 -G 1 g = o. 



In the transformation 0} P into P f , to a pencil oj rays corre- 
sponds a pencil oj conies projective to the pencil oj rays. The 
product oj the two pencils is a curve oj the third order. This 
curve may also be considered as the locus oj those points on the 
rays oj a pencil whose corresponding points are on the same rays, 
each for each. 

Ex. 1. Establish the equation of a coaxial system of circles. 
Prove the propositions of this section directly in this special case. 

Ex. 2. Prove that the pencil of rays joining any point to 
the centers of a coaxial system of circles is projective to this sys- 
tem. Establish the equation of the curve produced by the two 
pencils. 



THE STEIN ERI AN TRANSFORMATION. 185 

Ex. 3. Show in what manner a system of confocal conies 
may be considered as a range of conies. 

Ex. 4. What is the fundamental quadrilateral in case of two 
conies u = o, ^ = 0, having a double contact? 



§ 47. The Steinerian Transformation. 

1. In the foregoing sections we have shown that the polars 
of any point P with respect to a pencil of conies are concurrent 
at a point P' '. For the construction and clear understanding of 
this transformation it is of great advantage to consider in par- 
ticular the degenerate conies of the pencil through the quadrangle 
which shall be designated by A X A 2 A 3 A±, and its diagonal points by 
B ly B 2 , B 3 . The pairs of lines A X A 2 , B 3 A 3 ; A 2 A 3 , B X A X \ A 3 A U B 2 A 2 
are the degenerate conies of the pencil. To find P r when P is given, 
join P to B ±J B 2 , B 3 , Fig. 76, and construct the fourth harmonic 
rays to PB lf PB 2 , PB 3 with respect to the corresponding pairs 
of lines through B lt B 2 , B 3 . The three harmonic rays intersect 
each other at P'. From this simple geometric construction it 
is now easy to study the correspondence of P and P f for any 
particular positions. At every point B, say B 3 , the lines. A x A 2y 
B 3 A 3f B 3 B^ B 3 B 2 form a harmonic pencil. To the points B 
correspond, therefore, all points of their opposite sides of the tri- 
angle B X B 2 B 3 . The points A 1 ,A 2 ,A 3 ,A 4 are invariant, since the 
fourth harmonic rays pass through the points themselves. To 
a point on any line joining two of the fundamental points, say 
A t A 3 , corresponds the fourth harmonic point to the pair A^ 3 . 
All other points are in uniform correspondence. 

We have seen that to a straight line corresponds a conic. As 
a straight line cuts each of the sides B X B 2 , B 2 B 3 , B 3 B ly and as to 
these sides correspond the opposite points B 3 , B lf B 2 , it follows 
that said conic passes through the points B u B 2 , B 3 . To the 
straight lines of the plane corresponds the net of conies through 

1 See Steiner's collected works, Vol. I, pp. 407-421, and M. Disteli: Die 
Metrik der circidaren Curven dritter Ordnung im Zusammenhang mit geometrischen 
Lehrsatzen Jakob Steiners. Also Poncelet: Traite, 1 ed. 1822, p. 198. 



i86 



PROJECTIVE GEOMETRY. 



B X B 2 B 3 . Taking a pencil of rays through P, its corresponding 
pencil of conies passes through P' and B ly B 2 , B 3 . The curve 
of the third order produced by these two pencils passes, therefore, 
according to § 46, (4), through P and P', B u B 2 , B 3 . 

On every ray through P there are two corresponding points 
X and X' of the cubic. Consequently, connecting P to A l} A 2 , 




Fig. 76. 

A 3 , A 4 , the corresponding points on these four rays coincide, 
each for each, with A u A 2 , A 3 , A 4 , so that these points are on 
the cubic and PA t , PA 2 ,PA 3 , PA 4 the tangents at these points. 
On the rays PB U PB 2 , PB 3 the points which correspond to B x , B 2 , 
B 3 are the points of intersection 5/, B 2 ', B 3 r of these rays with the 
sides B 2 B 3 , B 3 B U B,B 2) respectively. The points J5/, B 2 ', B 3 ' are 
therefore also on the cubic. Hence the theorem: 



THE STEIXERIAN TRAXSFORMATIOX. 



187 



In the Steinerian transformation to every pencil oj rays corre- 
sponds a projective pencil of conies through the diagonal points 
oj the fundamental quadrilateral. The product of the two pencils 
is a curve of the third order through the vertex of the pencil of 
rays and its corresponding point and through the vertices and diag- 
onal points of the fundamental quadrangle. Thus to every point of 
the plane may be associated a certain curve of the third order in 
the Steinerian transformation. All these 00 2 cubics pass through 
seven fixed points. 

Without proceeding to the Steinerian transformation of 
conies, cubics, etc., we shall immediately take the general case 
of a cun r e of the wth order, C n - To determine in how many points 
any straight line g cuts C n , notice that the conic G corresponding 
to g cuts C n in 2n points. Hence, when the whole configura- 
tion is transformed, G with its 211 intersections on C n is trans- 
formed into 2n intersections of g with the transformed C n . Hence 
the theorem: 

In a Steinerian transformation a curve of the nth order is gen- 
erally transformed into a curve of order 2». 

2. Analytical Expression for a Steinerian Transformation. 

Nothing will be lost in the general result if we assume that 
the points A lt A 2i A 3 form an equi- A 2 
lateral triangle and that A 4 be its 
center, since by a collineation this 
orthogonal quadrangle may be trans- 
formed into any other quadrangle. 
Let A 4 coincide with the origin, and 
A x with the X-axis, Fig. 77, and 
A 1 A i =A 2 A 4 =A 3 A A =i. The foot- 
points of the perpendiculars of the 
triangle are B lt B 2 , B 3 . Consider 
first the degenerate conic consisting 
of the lines A X A 2 and A 3 B 3 with the 
equations 

c + V / 3-r / -i=o 

V3C- T) =0 




Fig. 77. 



1 88 PROJECTIVE GEOMETRY. 

so that the equation of the degenerate conic is 

(i) V3-l r2 -\ / 3-^+2^-v / 3-l : +^ = o. 

Similarly the equation of the degenerate conic represented by 
A t A 3 and A 2 B 2 is 

(2) X / 3'Z 2 -V~3'r) 2 -2ty-x / 3-^-7} = o. 

The equations of the polars of the point P (x, y) with respect 
to these conies are 



(3) (x\ / 3+y-W3)?+( x -y^ / 3+i) r )--^ / 3+-=o, 

(4) (xV^-y-W3)^-( x +y^3+h)y-7^-^ = °- 



The common solutions of (3) and (4) are the coordinates x*, y f 
of the point P corresponding to P in the Steinerian transformation : 

f ; 2(x 2 — y 2 ) + x 
I * = A {x 2 +f)-i> 

I ,_ y-ycy 
l y A (x 2 +y 2 )-i 



Solving these equations with respect to x and y we obtain 

r 

1 

(6) 1 



( 2(x' 2 -y' 2 ) + x' 

" 4 (V 2 -f/ 2 )-i ' 



y' — <\x'y' 

y 



4(x ,2 +/ 2 )-i ' 
which shows that the transformation is involutoric. 

y 

To the line at infinity, x =00, v=oo, — =arbitr., corresponds 
the circle 

(7) x' 2 +y' 2 = l. 



CURVES OF THE THIRD ORDER. 189 

Ex. i. The centers of the conies circumscribed to a quad- 
rangle A 1 A 2 A 3 A 4 lie on a conic K, which bisects the distances 
between these points in six points. These form three parallelo- 
grams having the same center, which is the center of the conic 
cutting out these points. 

Ex. 2. According as a straight g cuts K, in two real or two 
imaginary points, or touches it, the corresponding conic in the 
Steinerian transformation will be an hyperbola, an ellipse, or 
a parabola. 

Ex, 3. Prove by formulas (5) that in a Steinerian transforma- 
tion a C n is transformed into a C 2n . In particular a straight line 
is transformed into a conic. 

Ex. 4. Prove that in the Steinerian transformation A 1 A 2 A 3 A i 
are invariant points and that to the sides B X B 2 , B 2 B 3 , B 3 B 1 corre- 
spond the points B 3 , B 1} B 2 , by using formulas (5). 

§ 48. Curves of the Third Order. 

1. In the Steinerian transformation, with every point of the 
plane is associated a certain cubic. As in the previous section 
assume as conies determining the fundamental quadrangle or 
quadruple the degenerate conies. 

(1) u =v / 3-x 2 +2xy—\/? ) -y 2 —\ / 3-x-\-y = o, 

(2) u ± = V3 • x 2 — 2xy— V3 ■ v 2 — V3 • x— y = o. , 

To find the cubic associated with the point (V, /), take as 
lines g and g 1 in formula 8, § 46, 

(3) g = x-x' = o, 

(4) gi=y-y'=o. 

According to (5), § 47, to these lines correspond in the Stei- 
nerian transformation the conies 

(5) G = 2(x 2 -y 2 ) + x-x'\4(x 2 +y 2 )-i}=o, 

(6) G x = y-\xy -/{4(.v 2 +v 2 )-i|=o. 



19° PROJECTIVE GEOMETRY. 

The equation of the cubic associated with the point (V, /) 
is Gg 1 -G 1 g = o, or 

(y—y f ) { 2 (x 2 — y 2 ) + x— x'[a(x 2 -\- y 2 ) — i] ( 
-(x-x')\y-Axy - y '[ 4 ( x 2 + y 2 ) - i]j =o, 

or 

f J 2 {x 2 -y 2 ) + 
4(x 2 +y 2 ) 



(7) 



~-^j 



From the form of this equation it is apparent that a Steinerian 
transformation does not change the equation. Hence the theorem; 

The net of cubics through a quadruple and its diagonal points 
is invariant in the corresponding Steinerian transformation. 

This is also geometrically evident. In the construction of 
the curve, Fig. 76, eleven points are obtained through which 
the cubic passes and which, as a group, are invariant in the 
Steinerian transformation. 

y 

For the points oc f = ao, /=oo, —=/<:, (7) reduces to 



y—Axy 2(x 2 — y 2 ) + x 

(8) v+ kx+ \ V k ) 2 \\ =0. 

v J J A(x 2 + y 2 )— 1 4(x 2 +y 2 )— 1 



Also in this case the cubic is the locus of the double-points 
,of the involutions cut out on the pencil of parallel rays through 

the infinite point ( — = k j by the pencil of conies through the 

fundamental quadruple. The line at infinity belongs also to the 
pencil of parallel rays, and as the involution on it is rectangular 
it follows that the double points are the circular points. Hence 
(8) represents a pencil of bicircular cubics. 

As has been seen already, the tangents to the cubic at the points 
A t A 2 A 3 A 4 pass through the point P. We shall now prove that 
the tangents at B u B 2 , B 3 pass through P' . For this purpose 



CURVES OF THE THIRD ORDER. 



I 9 I 



draw a ray through P cutting the cubic in two points U and V, 
of which U shall be close to B 2 , Fig. 78. To this ray corresponds 




Fig. 78. 



in the Steinerian transformation a conic through B lt B 2 , B 3 , U, 
V, and P' '. As the ray through P turns in such a manner that U 
approaches B 2 as a limit, the corresponding conic will approach 
the degenerated conic, consisting of the ray P f B 2 and the side 
B x Bz as a limit. Hence, when the ray passes through B v the 
corresponding ray through P r will be a tangent to the cubic at B 2 . 
A similar result is obtained for the pointy B ± and B 2 , which proves 
the proposition. 



I 9 2 



PROJECTIVE GEOMETRY. 



2. In what follows it will be assumed that the cubic is a 
circular curve; i.e., that the point P is infinitely distant. Desig- 
nating this infinitely distant 
point by B and its correspond- 
ing point by C, the tangents at 
the A 's are parallel to the direc- 
tion of B, and the tangents at 
the B's pass through C, Fig. 79. 
The ray through C parallel to 
the direction of B is the asymp- 
tote of the curve. Hence the 
tangents at the points B, B t , B 2 , 
B 3 meet in the point C of the 
same curve. Four points on 
the cubic with this property 
are called a Steinerian quad- 
ruple of the cubic. 

Thus A X A 2 A 3 A±, BB X B 2 B 3 
are such quadruples. Accord- 
ing to previous results, the rays 
BB V BB 2 , BB 3 cut the opposite 
sides B 2 B 3 , B 3 B ly B X B 2 in three 
more points, C ly C 2 , C 3 , of the 
cubic. But this is equivalent 
with considering BB 1 B 2 B 3 as a 
fundamental quadrangle in a 
new Steinerian transformation 
with C 1} C 2 , C 3 as the diagonal 
points, and C as the original 
point associated with the cubic. 
That the cubic associated with 
C in this new transformation is 
identical with the original cubic 
follows from the following consideration: The points B being 
points of tangency count for eight given points. Furthermore, 
the four C's lie on the original curve, so that the new curve has 




Fig. 79. 



CURVES OF THE THIRD ORDER. 193 

at least twelve points in common with the original cubic, and is 
consequently identical with it. 

In this new Steinerian transformation construct the point D 
corresponding to C. Then take the new quadruple CC X C 2 C 3 and 
construct the associated cubic in the Steinerian transformation 
belonging to this quadruple. The new cubic is identical with 
the original cubic, as can easily be proved. The tangents at 
C, C lf C 2 , C 3 all pass through D. For the quadruple CC X C 2 C Z 
construct the diagonal points D lf D 2 , D 3 . These together with D 
form a new quadruple, whose tangents pass through E, the point 
corresponding to D in the transformation associated with the 
quadruple C x CJO£±. Continuing this construction, 1 we may 
obtain any number of points of the cubic arranged in quadruples. 
The points B, C, D, E . . . have the property that the tangent at 
one of these points always passes through the previous point. 

3. The general equation of a cubic may be written 

(1) Ax 3 +Bx 2 y+Cxy 2 +Dx 3 + 

ax 2 + 2bxy-\- cy 2 -\- 2dx+ 2ey-\-j = o. 

The problem arises, what connection exists between the 
fundamental quadruple with which the cubic is associated and 
the shape or equation of the cubic. In the above discussion the 
quadruple was assumed as real and the cubic consisted of a 
serpentine (infinite branch) and an oval. By certain collineations 
this curve may be transformed into various other curves which 
may be characterized with respect to their behavior at infinity. 
The serpentine or oval will be called elliptic, hyperbolic, or para- 
bolic, according as they have two imaginary, two real, or two 
coincident points at infinity. Designating by r the counter- axis 
which in a collineation is transformed to infinity, and by S and O 
the serpentine and oval of the cubic, then the transformed curves 
resulting from various positions of r are as given in the following 
table: 

1 For the sake of simplicity, in the figure only the quadruple CC X C 2 C 6 has been 
constructed. 



194 



PROJECTIVE GEOMETRY. 



r 


Original Curve. 


Resulting Curve. 


5 





5 





cutting 
tangent 
cutting 
tangent 


in 3 points 
in i point 




hyperbolic 
parabolic 
elliptic 
elliptic 


elliptic 
elliptic 
hyperbolic 
parabolic 




in 2 points 
in i point 







From these possible collineations it is seen that a cubic with 
two branches, serpentine and oval, by any collineation is trans- 
formed into a cubic with two branches. The geometrical dis- 
cussion of this section therefore does not cover all cases as repre- 
sented by the general equation of the cubic. For this purpose 
it is necessary to classify the cubics from the general equation, 
or the fundamental quadruple, by introducing coincident and 
imaginary elements. We shall do both. As the analytical dis- 
cussion is briefer, we shall take this up first and discuss the geomet- 
rical aspect later on. To equation (i) apply the general projec- 
tive transformation or collineation of the xy-plane as given in 
§ 19. This collineation depends upon eight parameters. After 
the transformation, clearing of fractions, collection of equal terms 
in x and y, (1) assumes the form 



(2) 



( A 1 x*+B 1 x 2 y + C 1 xy 2 +D x y*+ 
( a x x 2 + 2b 1 xy+c 1 y 2 + 2d 1 x+ 2e x y+ A 



where A lf B ly . . . a lf b lf . . . are polynomials in A, B, ... a,b, . . . 
and the eight parameters of the collineation. It is evidently 
possible to choose in an infinite number of ways the eight param- 
eters in such a manner that in (2) the coefficients B lf C\, D x , 
b u c x vanish, which amounts to five equations with eight un- 
known quantities. It is therefore possible to find a collineation 
transforming (1) into an equation of the form 



y 2 = ax 3 + fix 2 + yx + 



CURVES OF THE THIRD ORDER. 

or, resolving the right side into its linear factors, 



195 



(3) 



y i=a (x- e t ) (x- e 2 ) (x- e 3 ) 3 



in which e 1 has a different meaning from the e t used above. 

The general equation of the cubic can therefore always be 
reduced to an equation of the form (3), so that the discussion of 
the cubic with respect to its type may be limited to equation (3). 
This equation represents a curve 
which is symmetrical with respect 
to the x-axis, and its shape depends 
essentially upon the values of e lt 
e 2 , e s . Assume e^e^e^ The 
following cases may be dis- 
tinguished : 



--*+. 



I. e t 



are real and different 




Fig. 80. 



jrom each other. 

On the x-axis the curve has 
the real points with the abscissas 
e i> e 2, e 3> Fig. 80. In order that y 2 be positive, it is necessary that 

either e^x^-e 2 or x>e 3 . From this it 
follows easily that the cubic consists in 
this case of an oval and a serpentine. 
This is the case discussed in connection 
with the real quadruple. 

II. e x is real, e 2 and e 3 are conjugate 
imaginary. 

In this case we can write (3) in the 
form 

y 2 =a(x-e l )[(x-py+g 2 ], 
from which follows that y 2 is real only 
whenx>e x ; the curve consists of only 
This case is equivalent with a fundamental 
quadruple with two real and two conjugate imaginary points, as 
we shall see later on. 



- u +~- 




Fig. 81. 



one branch, Fig. 81. 



196 



PROJECTIVE GEOMETRY. 




Fig. 82. 



III. e 1 = e 2 and e 3 all real. 
Equation (3) assumes the form 

y 2 =a{x— e^) 2 (x— e 3 ). 

To get real values for y, x> e 3- 
The point x=e 19 y = o satisfies the 
equation also; but it is an isolated 
point, Fig. 82. 

Correspondingly, in the quadruple 
two points are real and two coincide. 

IV. e 1 and e 2 = e 3 are real. 



The equation becomes 

y 2 =a(x~e 1 )(x—e 2 )' 
y is real for x>e v Hence x= 



is a 



double -point of the cubic, Fig. 83. 

This case corresponds to a funda- 
mental 'quadruple with two coincident 
and two conjugate imaginary vertices. 



V. e x = e 2 



e 3 and all real. 




Fig. 83. 



Equation (3) can be written 

y 2 = (x-e 1 )\ 
We must take x>e v The curve 
has a cusp at x = e x with the rv-axis 
as a tangent, Fig. 84. The four points of the quadruple are 

real and three of them are coinci- 
dent along the tangent of the cusp. 
These are the five types of 
curves of the third order into 
which all cubics may be projected. 
Newton 1 called these five types, 
found by him, respectively, 

parabola campaniformis cum 
ovali, 

parabola pura, 
parabola puncta, 



n I e i e 2 e s 




Fig. 84. 



1 Eniimeratio linearum tertii ordinis (Londini, 1706). 



CURVES OF THE THIRD ORDER. 197 

parabola nodata, 

parabola cuspidata, 
and, according to their behavior at infinity, subdivided them 
into seventy- two different kinds. By later investigations six more 
were added to the seventy -two. 

As in the first case, this classification may be made by choos- 
ing in the perspective collineations the counter- axes r properly. 

§ 49. Curves of the Third Order Generated by Involutoric 

Pencils. 

1. Every straight line cuts a pencil of conies in an involution 
of points. Instead of any two conies <u = o t u 1 = o i we may take 
two degenerate conies with the same vertex : 

u =pp t =o\ 

«i = (P+*PdiP+PPi)=o, 

where p and p x represent two distinct straight lines. The pencil 
of conies then becomes 

u+ w^pp^ vKP+iptXP+fipO } =0, 

where v is a variable parameter. We may write this also in the 
form 

vp 2 + (1 + vp+ vX)pp 1 + vXixp x 2 = o. 

Solving for p, 

- (1+ vX+ vj J .)±\/( 1 + v2+v fl y- 4v *^ 

P= -A, 

and designating by £ and 7} the expressions multiplying p t in the 
last formula, the equation of the involutoric pencil of rays may 
be written 

(1) (P-$Pi)(P-VPi)=o, 

where ^y] = Xji = constant. 



198 PROJECTIVE GEOMETRY. 

For every set of values of £ and 77 satisfying this condition, (1) 
represents two rays of a pair of the involution 

p-rjp^o. 

The product of two projective pencils of this kind, having 
the same v, 

(2) ppi+A(P+ipi)(P+vPi)\=o, 

(3) Wi+ v{(q+ /^i)(?+ Ai) 1 =0, 

is evidently the curve of the fourth order: 

(4) ppi \ (q+ Hi) (q+ Mi) \-m\(P+ 4A) (P+ M) I =0, 

with the double-points p=p l = o and q = q l = o. 

In (1), p = o and p 1 =o are evidently the equations of a pair of 
the involution. 

In (3) the corresponding pair is given by q=o, ^ = 0. Letting 
the corresponding rays p t and q x coincide; i.e., p 1 = z q 1 = o i the 
curve (4) degenerates immediately into the ray p x =o and the 
cubic 

(5) p\(&+vpi) te+v-'Pi)\-q\(P+*Pi) (^+M)l=o. 

To distinguish the pencils (2) and (3) from ordinary linear 
involutoric pencils, we shall call them quadratic. The result 
may be stated in the theorem: 

The product of two projective quadratic involutions of rays 
is a curve of the fourth order. If the two involutions have two 
corresponding rays in common, then their product is a curve of 
the third order and that common ray. 

The cubic can also be produced by two projective pencils of 
which one is linear and one quadratic: 

(6) ( P+Vp " 



CURVES OF THE THIRD ORDER. 199 

whose product is 

(7) P\(q+ki) (?+/Wi)I-m?i=0' 

As (5) and (7) contain respectively twelve and ten arbitrary 
parameters it is clear that every cubic may be represented by 
one of these forms. 

2. Considering (5), each two pairs of corresponding rays of 
the two quadratic involutions (2) and (3), in which q 1 = p 11 



(8) 



\p-ypi=°, 



(9) 



q-?Pi=°9 
q-y'Pi=o, 



intersect each other in four points of the cubic. The vertices 
of the two pencils are also on the cubic. Two pairs of the quad- 
ratic involution in one pencil and the two corresponding 
pairs in the other pencil are therefore sufficient to determine 
the projectivity and consequently also the cubic, since they deter- 
mine ten points on the curve. 

Conversely, if on a cubic two vertices B and B x are known, 
and if it occurs twice that two rays through B cut certain two rays 
through B t in four points of the cubic, then these pairs determine 
two projective quadratic involutions of rays by which the entire 
cubic is produced. 

To prove this assume a ray, a, through B x passing very close 
to B. If the foregoing statement would not be true, the 
product of the involutions determined by the four pairs 
of rays through B and B 1 would be a curve of the fourth order, 
according to (4). On the ray a there would be two points 
cut out by the corresponding rays through B, which in general 
would be distinct. As the ray a in the limit approaches the 
ray through B l passing through B, these two points on a become 
coincident; i.e., B is a double-point of the curve (4). Similarly 
B x is also a double-point. A double-point on another curve 



200 PROJECTIVE GEOMETRY. 

counts for two points of intersection, so that the supposed curve 
of the fourth order has twelve points in common with the given 
cubic. Construct the net of quartics (curves of the fourth order) 
through these twelve points. Any two points different from these 
twelve points, with these, determine fourteen points; i.e., such 
a quartic and only one, which therefore consists of the given cubic 
and the straight line through the assumed two points. 

From this it follows that there is only one quartic through 
the twelve points having B and B 1 as double-points, and this 
consists of the given cubic and the line through B and B u 
which proves the proposition. 

3. Consider three quadratic involutions of pencils of rays 
projective to each other: 

(10) PPi+M(P+tyi)(P+tfi)} =°> 

(11) qqi+v\(q+^qi)(q+v'qi)\ =0, 

(12) rr 1 +v{(r+X"r 1 )(r+ n"^)} =o, 

and suppose that the product of (10) and (n) is identical with 
the product of (10) and (12) ; i.e., that the two equations 

(13) ppA(q+ ^1X0+ fa) i- I (P+ *Pi)(P+ M) toi =0, 

(14) MlH^i)HA))-i»+«(^^i) K=o, 

must be simultaneously satisfied for all sets of values of x and y. 
This can only be true if 

(15) ^ 1 i(^^ 1 )(r+/v 1 )p^ 1 )(?+^ 1 )(?+^ 1 )!=o 

simultaneously with (13) and (14)- 

From this the theorem follows: 

// a quadratic involution 0) rays produces with two projective 
quadratic involutions one and the same quartic or cubic, then the 
product of the last two involutions is the same quartic or cubic. 



CURVES OF THE THIRD ORDER. 



201 



4- Considering again the construction of a cubic by the Stein- 
erian transformation, Fig. 85, and taking B at an infinite distance 




Fig. 85. 



in the indicated direction, then to a point X on the cubic corresponds 
a point Y on a ray through X parallel to this direction. Joining 



202 PROJECTIVE GEOMETRY. 

X and Y with B 2 and producing XB 2 and YB 2 to their intersec- 
tions Y x and X x with the cubic, then X± corresponds to F x in the 
Steinerian transformation and hence X x Y t is parallel to XY. 
It is now clear that the pairs of rays XY ly X t F through B 2 and 
XY, X 1 Y 1 through B, furthermore the pairs A,A 3 (counted 
twice) through B 2 and BA l7 BA 3 , determine two projective quad- 
ratic involutions around B 2 and B whose product is a cubic, 
since they have the ray B 2 B in common. This cubic having 
ten points in common with the cubic of the Steinerian trans- 
formation (A lf A 3 ; each counted twice, since BA X , BA 3 are tan- 
gents at A t and A 3 \ B, B 2 , X } Y, X l5 F 1 ) is identical with it. If 
we connect X and Y with B 3 and produce XB 3 and YB 3 to their 
intersections X/ and F/ with the cubic, then X/ and F/ cor- 
respond to each other in the Steinerian transformation; i.e., 
X/F/||XF. Consequently the cubic may also be considered 
as the product of the involution around B and an involution 
around B 3 . In the same manner it is also the product of the 
involutions' around B and B x ; hence, according to the foregoing 
theorem, the cubic is also the product of the two involutions 
around B 3 and B v Hence the points where X x B y and YJ$ X pro- 
duced meet the cubic are the same as X/, F/. In a similar 
manner it can be proved that XB X and X X B 3 , YB L and Y X B 3 
intersect each other in the points X r , Y r of the cubic, so that 
X'Y f \\XY. X has been assumed as any point of the cubic, and 
X x in such a manner that the corresponding point F of X lies 
in a straight line with X 1 and B 2 . Consider now the pairs of 
rays XX 1 , ,XB 1 and X t X r , X,B 3 ; and XY, XY, and X.Y,, XJT; 
they determine two projective quadratic involutions about X 
and X x whose product is a cubic which is identical with the original 
cubic, since it has ten points in common with it. Taking any 
point G on the cubic * and letting XG and Xfi cut the cubic in 
/ and K, then XJ and XK produced cut the cubic in one and 
the same point H; XG, XH and X X G, X t H form two corre- 
sponding pairs of the involutions around X and X v If G ap- 
proaches the point of intersection of XX X with the cubic, then / 

1 For the sake of simplicity in the figure the following part of the construction 
is not shown. 



CURVES OF THE THIRD ORDER. 203 

and K approach X x and X ; hence, in the limit, the tangents to 
the cubic at X and X t intersect each other in a point of the cubic. 
In a similar manner it can be proved that the tangents at Y and 
Y lt X' and X x ' , Y' .and Y ( intersect each other in points of the 
cubic. Again, G and H, and / and K may be assumed as 
vertices of projective quadratic involutions producing the cubic. 
Hence also the tangents at G and H, and / and K intersect 
each other in points of the cubic. We can therefore state the 
following theorem : 

Designating two points on a cubic whose tangents at those 
points intersect each other in a point of the cubic as a Steinerian 
couple, or simply as a couple, then the cubic can be produced by 
two projective quadratic involutions around these points. 

The lines joining any point oj the cubic to the points of a couple 
cut the cubic again in a couple, and all couples of the cubic are 
produced when this point describes the whole cubic. 

Each two corresponding pairs of the involutions around the 
two points 0} a couple intersect each other in two new couples. 
Such two involutions produce all couples of the cubic. 

A quadruple on a cubic is defined as a- group of four points 
any two of which form a couple; i.e., the tangents at the four 
points concur in a point of the cubic. From this definition we 
infer easily: 

The lines joining four points of a quadruple cut the cubic in 
another quadruple. The sixteen lines joining the points of two 
quadruples intersect each other, four by four, in four points of a 
new quadruple. 

These results form a part of the theory of problems of clo- 
sure on the cubic as it has been developed by Steiner, Clebsch, 
and others. 1 They are sufficient for the applications in the 
following sections. 

1 For further details references are made to 

Clebsch: Crelle's Journal, Vol. LXIII, pp. 94-121. 

Steiner: Crelle's Journal, Vol. XXXII, pp. 371-373. 

Disteli : Die Steiner 'schen Schliessungsprobleme nach darstellend-geometrischer 
Methode. Leipzig, 1888. 

Emch: Applications o{ Elliptic Functions to Problems of Closure, University 
of Colorado Studies, Vol. I, pp. 81-133. 



204 PROJECTIVE GEOMETRY. 

Ex. i. Verify the theorems of this section constructively, 
when B is finite or infinite. 

Ex. 2. What relation must exist between a quadratic and a 
projective linear involution of rays in order that the cubic pro- 
duced be one with a cusp. 

Ex. 3. Prove directly that a cubic can be produced by two 
quadratic involutions around the points of a couple by determin- 
ing two corresponding pairs of the involutions. 



§ 50. Various Methods of Generating a Circular Cubic. 

1. In § 48 (8), we found for the equation of the bicircular 
cubic referred to the equilateral triangle A X A A 9 with A A as 
point of concurrence of altitudes, after some rearrangement, 

(1) 4(kx— y) (x 2 + v 2 ) — 2 kx 2 — 4xy+ y+ 2y 2 — 2 kx+ 2V = 0. 

The slopes of the asymptotes at the circular points are evi- 
dently -\-i and —i, so that the equations of these asymptotes 
are of the form 



where c x and c 2 are constants to be determined. If equations (2) 
represent asymptotes, then their common solutions with (1) 
must give infinite values for x and y. Substituting the values 
of y from (2) in (1), we get respectively 

J (Sc^ 8c. xi— 4i— 4k)x 2 + Bx-\- C =q, 
^ 3 ^ { (8c 2 - 8c 2 ri+ 41- 4K)x 2 +B f x+ O =0, 

where B, C\ B', C are polynomials in c v k; c 2 , k, different 

from those of the x 2 's. 

In both cases the values of x will be infinite, if we have re- 
spectively 

1 K+i 

8c, + 8c { Ki—4i—4K=o, or c,= ■ — :, 

11 ^ ' 2 1 + Kl 



CIRCULAR CUBIC S. 



205 



and 



8c 2 — Sc 2 Kt+^i— 4K=o i 



1 k— 1 
or c, = :, 

2 i-«r 



so that the equations of the asymptotes are 



(4) 



I K+t 

y = tX-\ ; :, 

J 2 1+ Kl 



y 



1 K—l 

tX+ :. 

2 I— Kl 



Solving these two equations simultaneously, we get for the 
coordinates of the point of intersection of the two asymptotes (4) 



(5) 



1 /r— 1 k 

JV+l' y= K 2 +I 



The sum of the squares of these expressions is x 2 +y 2 = j] i.e. f 
the point of intersection is on the circle corresponding to the 
infinite line in the Steinerian transformation. The real asymp- 
tote of the cubic has the slope k, so that its equation is of the 
form y=KX+c 3 . By a similar method as in the case of c x and 
c 2 above, we find 

3#C— K 3 

C3 = T+2~?' 

and as the equation of the real asymptote 



(6) 



y= kx+ 



3«~* 

I + 2K 2 ' 



Solving (1) and (6) simultaneously, we find for the point of 
intersection of this asymptote with the cubic 



(?) 



x=- 



1 I — K- 

2 I+/C 2 ' 



V= — 



I- 1 - K' 



206 PROJECTIVE GEOMETRY. 

Comparing (7) with (5), it is seen that the two points are 
diametral. 

A similar result is obtained by taking any orthogonal quad- 
ruple A X A 2 A^A^ and the circular cubic associated with it. In 
this case the equation of the cubic assumes the form 

(8) (ax+@y)(x 2 +y 2 )-\-ax 2 +2bxy+cy 2 +2dx+2ey J r}=o. 

Repeating on this equation the same process as above on equa- 
tion (1), the theorem may be stated thus: 

Considering a bicircular cubic in a Steinerian trans formation, 
the asymptotes at the circular points intersect each other in a point 
D of the circle which corresponds to the line at infinity in the Stei- 
nerian transformation. The real asymptote cuts the same circle 
in a point C which with D determines a diameter of the circle. The 
points D and C are called center and principal points of the cubic. 

2. In equation (8) the infinitely distant real point of the cubic 
is the infinite point of the line ax-\-fty = o, as can easily be verified. 
Taking the x-axis parallel to this line, (8) becomes 

(9) y{oc 2j r y 2 ) + ax 2 -\- 2bxy-\- cy 2 + 2dx+ 2ey+f=o. 

In a similar manner as in (5), the coordinates of the center of 
the cubic are found to be 

, a—c 

(10) x=b, y=—^-. 

Taking this point as the origin of a new coordinate system 
with axes parallel to those in (9) , (9) assumes the form 

(11) (y+a)(x 2 +y 2 ) + 2dx+2ey+f = o. 

Here the equation of the real asymptote is y = — a, so that 
the coordinates of the principal point of the cubic become 

2ae— f 



CIRCULAR CUBICS. 207 

Equation (11) may be considered as the result of the elimina- 
tion of P from the two equations 

x 2 + y 2 -P = o, 
2dx+ 2ey+)+ P(y+ a) =0, 

which represent two projective pencils of concentric circles and 
rays. Hence the theorem of Czuber: 1 

Every circular cubic may be generated by two projective pencils 
oj concentric circles and rays. The common center of all circles 
of the pencil is the center of the cubic, and the vertex of the pencil 
of rays is the principal point of the cubic. 

3- In § 49 it has been shown that the points on a cubic may 
be arranged according to couples, so that the rays from any 
point on the cubic to the points of these couples form an involu- 
tion. 

Suppose now that the direction of the real asymptote of a 
circular cubic is perpendicular to one of the sides, say B 2 B 3 of 
the diagonal triangle B X B 2 B Z , Fig. 86; then the center of the 
cubic will coincide with the point B t . In other words, the cir- 
cular points form a couple, so that the involutions of rays from 
any point of the cubic contain the directions of the circular points 
as a pair. 

Hence, according to a theorem in § 5, p. 21, since A t A 3 , A 2 A 4 
are two couples and P any point on the cubic, the angles A X PA Z 
and A 2 PA± are equal. Hence the theorem: 

The circular cubic which contains the circular points as a 
couple (conjugate pair) is the locus of all points from which two 
fixed lines A X A 3 , A 2 A 4 appear under the same angle. 

Inscribing a conic to the quadrilateral A X A 2 A 3 A 4 , then the 
pieces A t A 3 and A 2 A 4 of the tangents, contained between the two 
other tangents A X A 21 A 3 A 4 of the conic, are subtended by equal 
angles at the focus, § 35, p. 118. Hence the theorem: 

1 Die Kurven dritter und vierter Ordnung, welche durch die unendlich fernen 
Kreispunkte gehen. (Zeitschr. f. Math., XXXII, 1887.) 



208 



PROJECTIVE GEOMETRY. 



The locus of the Joci o] all conies inscribed to a quadrilateral 
is a circular cubic having the circular points as a conjugate pair. 

The same cubic may also be produced by two projective pencils 
of circles over A X A 3 and A 2 A^ in which two corresponding circles 
subtend equal peripheral angles over the chords A X A 3 and A 2 A 4 . 




Fig. 86. 



But if two projective pencils of circles G+ XG t = o and G' ' + XG x f 
= o produce a cubic, say GG 1 f —G 1 G f =o, so that this equation 
reduces to Gg/—G x g , = o ) where g/ and g' are linear factors, 
then the cubic may also be produced by two projective pencils 
of circles and rays. 

4- In the same circular cubic consider the pencil of circles 
through B 2 B 3 , Fig. 86. The ray Bfi passes through the center 
of the circle through B X B 2 B Z . A 2 A 3 passes through the center 
of the circle through A 2 A b (diameter of said circle) and B 2 B 3 . 
B r A^A x passes through the center of the circle through A A A X 
(diameter) and B 2 B 3 . The three circles through B 2 B 3 and the 



CIRCULAR CUBICS. 209 

three corresponding rays through B x determine nine points of the 
cubic, since B x as a point of tangency on B X C counts twice. The 
two pencils therefore determine two projective pencils of circles 
and rays whose product is the given cubic. Hence the theorem: 

The circular cubic having the circular points as a conjugate 
pair is also the product 0) a pencil of circles and a projective pencil 
of rays which pass through the centers of the corresponding circles. 

Ex. 1. With the Steinerian transformation for base, prove that 
the general equation of a circular cubic has the form 

(ax-\-Py)(x 2 +y 2 ) + ax 2 4-2bxy+cy 2 -\-2dx+2ey J r j = o. 

Ex. 2. Given the pencil of circles 

x 2 +y 2 — p 2 — 2Ax = o (p = constant) 

and a pencil of rays passing through the centers of these circles. 
To find the equation of the product of the two pencils and discuss 
the result for different positions of the vertex of the pencil. 

Ex. 3. Establish the equations of two projective pencils of 
circles in which corresponding circles subtend equal peripheral 
angles over the fixed chords. 

Ex. 4. Prove that in a circular cubic the oval and the ser- 
pentine appear under the same angle from any point of the curve. 

Ex. 5. The extremities of two diameters A X A 2 and A 3 A 4 
form a square. 'What is the locus of the points from which both 
diameters appear under the same angle? 

§ 51. The Five Types of Cubics in the Steinerian Transfor- 
mation. 1 

I. Cubic with Oval and Serpentine. 

This cubic is obtained when all four points of the funda- 
mental quadruple are either real or imaginary. As the case 

1 This section has been published in The University of Colorado Studies, 
Vol. I., No. 4, Feb. 1904. 



2IO 



PROJECTIVE GEOMETRY. 



with four real points has so far always been used to illustrate the 
general properties, we shall now assume an entirely imaginary 
quadruple determined by a coaxial system of circles with the 
limiting points P and Q, Fig. 87. On every ray g through an 




Fig. 87. 

arbitrary fixed point B the circles of this system cut out an invo- 
lution of points whose double-points X and X' are two points of 
the cubic associated with the point B in the Steinerian transfor- 
mation belonging to the given imaginary quadruple. To construct 
X and X', let g cut the line m, which is the line joining the finite 
imaginary points of the quadruple, at M. With M as a center 
pass a circle K through P and Q which will cut g in the required 



CIRCULAR CUBICS. 211 

points. In reality, according to the well-known construction 
just explained X and X' are the points of tangency of g with two 
circles of the given system. It will be noticed from the figure 
that the two points of the cubic on a ray through B are always 
equally distant from m. Hence, taking a ray through B parallel 
to w, the point at infinity corresponding to Q will be in a line 
through P parallel to m. In other words, the line through P 
parallel to m is the asymptote of the cubic.^Xonsideri»g^the 
pencil of circles through P and (X^the same circular cubic is also 
produced by this pencil and^the pencil of diameters through B. 
Thus a statement in the foregoing section is corroborated. 

II. The Simple Cubic. 

This curve is produced by assuming two separate real and 
two imaginary points in the fundamental quadruple. In Fig. 87 
let P and Q be the real points, and the circular points of the pencil 
of circles through P and Q the imaginary points. To find the 
points Y and Y r where a ray / through B cuts the cubic, let / 
cut n at N. With N as a center construct the circle L orthogonal 
to the pencil of circles through P and Q. The circle L cuts / 
in the required points Y and Y' . This cubic appears again 
plainly as the product of a pencil of circles and the pencil of 
diameters through B. Two points on a ray through B, like 
Y and Y' y are always equally distant from n. To R corresponds 
the infinitely distant point of the cubic ; consequently, the asymp- 
tote is parallel to n and its distance SC from n is equal to RC. 

Ex. 1. Prove that the two cubics in Fig. 87 intersect each 
other orthogonally. 

Ex. 2. Construct the tangents to the two cubics at B, P, Q. 

III. Cubic with an Isolated Point. 

The quadruple consists of two distinct points A lt A 3 and two 
coincident points A 2 , A 4 . It is assumed that the direction of 
the line joining A 2 and ^4 4 in the limit; i.e., as they become coinci- 
dent, cuts A t A 3 at B 2 . B t and B 3 will coincide with A 2 and A it 
Fig. 88. Joining B, which, as usual, has been assumed infinitely 
distant, to B lt B 2 , B 3 , and constructing the fourth harmonic rays 
to the pairs of sides passing through these points, it is seen by 



212 PROJECTIVE GEOMETRY. 

passing over to the limit that the fourth harmonic rays at B x and 
2*3 coincide. As before, they cut the fourth harmonic ray through 
B 2 at C, the point through which the asymptote passes. 

The pencil of conies through the quadruple cuts every ray 
through B to the left of A 3 and to the right of A x in elliptic in- 





A 


1/ 


b$/r 


/ 


/ 


V / 


/ 


/ 


/ V 


/ 


/ 


yi/. / p^ 








/BjB. 










/ &"' / / 






/ A i / ' 












/ 






? x 






Fig. 88. 







volutions, and only the rays between A x and A 3 contain hyperbolic 
involutions. The only branch of the cubic is therefore contained 
between two lines through A x and A 3 parallel to the direction 
of B. The ray through A 2 A 4 carries a parabolic involution, and 
A 2 A 4 represents an isolated point of the cubic. 

IV. Nodal Cubic. 

Assuming in the fundamental qnadruple A x and A 4 real and 
coincident and A 2 , A 3 conjugate imaginary, a cubic with a double- 
point or node at A X A 4 is obtained. In Fig. 89 a vertical line 
through A X A 4 represents the limiting direction of the line joining 
the two points. As conies of the pencil through the fundamental 
quadruple take the pencil of circles tangent to each other at A t A 4 
and to the vertical line. A 2 and A 3 are then represented by the 
circular points at infinity. To construct the cubic associated with 
an arbitrary point B, draw rays through B. On each of these 
rays the pencil of circles cuts out an involution whose double- 



CIRCULAR CUBICS. 



213 



points are points of the cubic. These points are, of course, also 
the points of tangency of circles of the pencil. Hence, to find 
the points where a ray g through B cuts the cubic, take the point 
M where g cuts m, the line joining A x with A 4 , as a center of a 
circle K passing through A ± A A . K cuts g in the required points 
X and X f . From this it is seen that the nodal cubic is also 




the product of a pencil of circles with coincident limiting points 
and a pencil of diameters. As X and X' are equally distant from 
m, the asymptote is parallel to m at a distance to the left of m 
equal to BA X (BA^ _L m for the sake of symmetry.) 

Y. Cuspidal Cubic. 

In this case three of the four real points of the fundamental 
quadruple coincide. Constructively this can be arranged best by 
assuming as the pencil of conies a pencil through a fixed point A t 
and with its conies all osculating each other at another fixed 
point which evidently may be considered as the representative 
of the three coincident points A 2 , A 3 , A 4 . 

To construct a pencil of osculating conies we may start 
with the construction of § 41, 9, Fig. 67. There it was shown 
that the picture of a circle in a perspective collineation whose 



214 



PROJECTIVE GEOMETRY. 



center lies on the axis of collineation and also on the given circle 
is a conic osculating the given circle at the center of collinea- 
tion. Hence, considering in Fig. 90 the line 5 joining A x 
with the coincident remaining points as the common axis of an 
infinite number of perspective collineations in which only the 





fsf* 



>(W, G, / / / a'. 







Fig. 90. 



counter-axes vary, the pictures of a fixed circle K through 
A X A 2 A 3 A 4 form clearly a pencil of osculating conies. 

On every ray g' (or the identical g x ') through a fixed point B 
(assumed infinitely distant) this pencil cuts out an involution 
whose double-points are two points on the cuspidal cubic asso- 
ciated with B in the Steinerian transformation. These points 
are also the points of tangency of g f (g/) with two conies of the 
pencil. For the actual construction of these points the following 



CIRCULAR CUBICS. 215 

simple method may be applied: Let g' intersect s at S. From 
S draw the two tangents g and g x to the circle K. 1 Through the 
center of collineation or the cusp draw a line / parallel to the 
direction of B. Let T and 7\ be the points of intersection of I 
with g and g lf and through T and 7\ draw two lines r and r x 
parallel to s. Considering r and r x as counter-axes of two dis- 
tinct collineations with the same axis s and the same center, then, 
according to the constructions of collineations, g r and g/ will be 
the pictures of g and g ± in these two collineations, and the rays 
joining A to G and G t cut g' (g/) in two points G' and G/ which 
evidently are the points of tangency with g' (g/) of the two oscu- 
lating conies corresponding to the fixed circle K in the two colline- 
ations (r, i\). The line I cuts K at U; the tangent at U cuts 5 
at V] and from the construction follows that the line through V 
parallel to I is an asymptote. In a similar manner the lines join- 
ing C to the points of tangency W and W ± of the tangents to K 
parallel to 5 are the directions of the two other real asymptotes. 
By a suitable collineation this cuspidal cubic may be transformed 
into the symmetrical form of Newton's parabola cuspidata. 

Ex. 1. Prove that if s is a diameter of K and the direction 
of B is perpendicular to s, then the above cubic degenerates into 
an equilateral hyperbola. 

Ex. 2. Prove that if K is tangent to s, then the cubic degener- 
ates into a parabola. 

Ex. 3. A pencil of cubics is determined by two cubics or by 
eight arbitrary points of which no four are in the same straight 
line. But it is clear that the two cubics determining the pencil 
have nine points in common, hence all cubics of the pencil pass 
through these nine points. In other words: All cubics passing 
through eight fixed points pass through a ninth fixed point. 

Ex. 4. Through four points A, B, C, D of a cubic draw the 
pencil of conies (K). Every conic K of this pencil cuts the 
cubic in two points P and Q. Prove that the secant PQ cuts 
the cubic in a fixed point. 

1 In Fig. 90 K is the only circle shown, and / is the line through A cutting 
this circle at U. 



216 PROJECTIVE GEOMETRY. 

Ex. 5. Let two straight lines / and m cut a cubic in the points 
A, B, C and D, E, F. Construct the points of intersection 
G, H, I of AD, BE } CF with the cubic and prove that they are 
collinear. 

Ex. 6. Construct the cubic in the Steinerian transformation 
when one of the points of the quadruple is infinitely distanto 



CHAPTER V. 

APPLICATIONS IN MECHANICS. 

§ 52. A Problem in Graphic Statics. 

1. Let 1, 2, 3, ... be a system of coplanar forces in a plane, 
Fig. gia. With O and O' as poles construct two force-polygons, 
Fig. 91&, and in the previous figure the two corresponding funicu- 
lar polygons. Considering in both figures the lines o, 012, o'i2, 
o', 12, it is seen that corresponding lines are parallel and that 
they form in each case five sides of a complete quadrilateral. 
Hence, according to the last theorem of § 25, also the line joining 
the intersections of o and o', and of 012 and o'i2, in Fig. 91a, is 
parallel to 00' in Fig. gib. In a similar way it can be proved 
that the line joining the intersections of 012 and o'i2, and of 0123 
and o'i23, in Fig. 91a, is parallel to 00' in Fig. 916, i.e., identical 
with the line joining (o and o') with (012 and c/12). This result 
may evidently be extended to any number of^ forces, so that we 
have the theorem: 

Corresponding sides of two funicular polygons of a system 
of coplanar forces intersect each other in points of the same straight 
line. 1 

Corollary. — // the forces are concurrent, they and the two 
funicular polygons determine a perspective collineation. 

2. The value of this theorem will appear from the solution 
of the following problem: 

Two bars, AC and BC, connected by a pivot- point at C, are 
supported by pivots at A and B (Fig. 92). Two forces, 1 and 2, 

1 In Cremona's Graphic Statics this theorem follows from the fact that the two 
figures (a and b) form two reciprocal figures. See loc. cit., p. 127. 

217 



21, 



PROJECTIVE GEOMETRY. 



are applied to the bar AC, and in the same manner two forces, 
j and 4, to the bar BC. To find the reactions at the points A, B,C. 




Fig. 91a. 




Fig. 91&. 



First determine magnitude, direction, and position of the 
resultants (12) and (34) of the forces 1, 2 and 3, 4 by means of 
the polygon of forces (Fig. 926) and the funicular polygon 
(Fig. 92a). Then construct the funicular polygon of the result- 



APPLICATIONS IX MECHANICS. 



219 



ants (12) and (34) with O' as a pole and with its first side pass- 
ing through A. Every funicular polygon constructed in this 
manner will be collinear with ever) 7 other, and with the point 
of intersection M of (12) and (34) as the center of perspective 




Fig. 92a. 



collineation. Now it is clear that the polygon passing through 
A , C, and B, and formed by the reactions at these points, is also a 
funicular. It is therefore collinear with the first polygon (o', o'i2, 
c/1234). Projecting the points C and B from M upon the funicu- 
lar sides (c/12) and ((/1234), respectively, the projected points 
C and W will correspond to C and B in a perspective collinea- 
tion. Hence the fines BC and B'C will intersect each other 



220 



PROJECTIVE GEOMETRY. 



in a point S of the perspective axis s. 
point A the axis is determined. 



By this point and the 




The directions of the reactions at A, C, and B intersect the 
funicular lines o', o'i2, ©'1234 in points of the line s, and they 
may easily be drawn. To find the magnitudes of the reactions, 
draw lines parallel to their directions through the points A u C lt 
B v These lines necessarily meet in a point 1 of the straight 
line g f . Thus O x A xy O x B Xi O l C l are the magnitudes of the reac- 
tions at the points A , B } C. 



§ 53. Statical Proofs of Some Projective Theorems. 

1. Constructing a funicular polygon of a system of coplanar 
forces 1, 2, 3, . . . , n, it is known that the resultant of the system 
passes through the point of intersection of the forces (o) and 
(0123...W) of the funicular polygon, and is also the resultant 



APPLICATIONS IX MECHANICS. 



221 



of these two extreme forces, with (o) reversed. Hence, when 

the system is in equilibrium, the forces (o) and (0123 ri) 

must coincide. Hence the theorem: 

A funicular polygon of a system of coplanar forces in equilibrium 
is a closed figure. 

Consider now three forces 1, 2, 3 in equilibrium, Fig. 93a, and 
draw any triangle o, 01, 012 having its vertices on these forces. 
Draw also, in Fig. 936, the force-polygon 123. Through the inter- 




Fig. 93k 



Fig. 93a. 



section of 1 and 3 in (b) draw a line parallel to o in (a) ; through 

1 and 2 one parallel to ci in (a). These two lines intersect each 
other in a point O. Now connect O with the intersection of 

2 and 3. Thus three forces OA, OB, OC are obtained, and 
if O is assumed as a pole and starting out with the original line 
o in (b), a funicular polygon is obtained which coincides with the 
original triangle o, 01, 012. Hence the theorem: 

Any triangle whose vertices lie on the lines of action of three 
forces in equilibrium may be considered as a funicular polygon 
of these forces. 

Consequently, according to the theorem of § 51, if we take 
any two triangles with their vertices on the three lines of action, 




222 PROJECTIVE GEOMETRY. 

the three points of intersection of corresponding sides are collinear. 
As any three concurrent lines may be chosen as lines of action 
of three forces in equilibrium, we thus have proved the well- 
known theorem concerning homologous triangles. 

2. Theorem. The middle points of the diagonals of any com- 
plete quadrilateral are collinear. 1 

Minchin 2 gives the following proof for this proposition : Let 
ABCDEF be the complete quadrilateral. Take the following 

system of forces, supposed act- 
ing on a rigid body: two forces 
represented by DA and DO in 
" magnitudes and senses, and 
two represented by BA and BO, 
Fig. 94 

Now the resultant of the 
first pair passes through a, the 
middle point of AC; so does 
FlG - 94- the resultant of the second 

pair; therefore the resultant of the four forces passes through a. 
Also the resultant of DA and BA passes through /?, the middle 
point of BD; so does the resultant of DC and BO; hence the 
resultant of the four forces also passes through /?. We shall now 
show that it passes through y, the middle point of EF. For 
this purpose introduce a force ED and a force DE which do not 
alter the given system. Introduce also forces CE, EC; CF, FC; 
FB, BF. Hence the given system is equivalent to forces EA, 
AE; DF, DE; BE, BF; EC, FC; and it is obvious that the 
resultant of each of these pairs passes through j; hence the 
resultant of the whole system passes through y. Now as the 
resultant of the given system acts in a right line, and as a, /?, y 
have been independently shown to be points on this resultant, 
these points are collinear. q.e.d. 

3. Pascal's theorem, that the intersections of the opposite 



1 Chasles, loc. cit., arts. 344, 345. 

* Treatise on Statics, Vol. I, pp. 145, 146. 



APPLICATIONS IN MECHANICS. 



223 




sides of a hexagon inscribed in a circle lie in a right line, is easily 
exhibited as a case of the 
funicular property in § 52. 
Following again Minchin, loc. 
cit., let the lines DA, EB, FC 
in Fig. 95 be lines of action 
of three forces, P, Q, R, such 
that if P is resolved at A into 
two components along AB, 
AF, or into two at D along 
DC, DE ; if Q is resolved into 
two at B along BA, BC, or 
into two at E along ED, EF; 
and if R is resolved at C 
along CB, CD, or at F along / 
FE, FA, the two compo- 
nents thus obtained along 

any side are equal and opposite. Obviously such conditions are 
consistent, on account of the equality of angles in the same seg- 
ment of a circle. Now if P, Q, R are applied at A, B, C, by 
the nature of the case a polygon FA BCD of jointed bars pivoted 
at F and D would be kept in equilibrium; i.e., this is a funicu- 
lar of the forces. Again, let P, Q, R be applied at D, £, F to 
a polygon CDEFA of jointed bars pivoted at C and A. This 
polygon would be in equilibrium, and a funicular of the forces. 
The two funiculars of the same forces, however, have the property 
that the intersections, a, /?, y, of their corresponding sides (AB, DE), 
(BC, EF), (CD, FA) are collinear. q.e.d. 

Ex. Prove that the medians of a triangle are concurrent. 



Fig. 95. 



Geometry of Stresses in a Plane. 



§ 54. General Remarks. 

Forces acting on a body cause certain displacements or strains 
between its particles. These strains are said to be within the 



224 PROJECTIVE GEOMETRY. 

elastic limit if after the disappearance of these forces the strains 
disappear also; i.e., if the body returns to its original condition. 
The forces which occur within the body as a result of the strains 
are called stresses. These are called tensions, compressions, or 
shears, according as their tendency is to pull the particles apart, 
to press them together, or to push them over one another. Accord- 
ing to Hooke's law the stresses in a body are approximately pro- 
portional to the corresponding strains as long as they occur within 
the elastic limit. In many cases, plane surfaces may be passed 
through strained bodies orthogonally to which there are no 
strains and consequently no stresses. This is the case in beams 
under tension, compression, or bending, and covers a great num- 
ber of engineering structures. In these cases the investigation 
of strains and stresses is limited to the plane. In what follows 
only stresses in a plane will be considered. 

The forces producing the stresses in a body and these them- 
selves are in equilibrium. The stresses in any portion of the 
solid are also in equilibrium. Considering an infinitesimal 
plane section in a strained solid, we make the assumption that 
the stresses acting on this element are uniformly distributed, 
so that their resultant passes through the center of gravity of 
this element. For many purposes it is convenient to consider 
the resultant stress per unit of the surface-element. This stress, 
the resultant divided by the element, is called the specific stress 
acting on that element. 

§ 55. Involution of Conjugate Sections and Stresses. 

1. Calling a plane surface through a strained body with the 
stresses acting in this plane (no stresses normal to the plane, as 
assumed above) a field of forces, we assume that under the influ- 
ence of this field every portion of this plane is in equilibrium. 
Thus, if a very small triangle ABC (infinitesimal in all rigor) 
is cut out of the field, the resultant stresses acting upon its sides 
must be in equilibrium. According to the assumption of the 
uniform distribution of stresses over an infinitesimal section, 



APPLICATIONS IN MECHANICS. 



225 



these resultants must pass through the middle points 7*, a, /? of 
the sides AB = c, BC = a, CA=b, and, being in equilibrium, are 
concurrent. Designate these resultants by A, B, C, as shown 
in Fig. 96. Each two of these forces, for instance A and B, may 
be resolved into components parallel to the sides AC and BC. 
Let A lf A 2 and B L , B 2 be these components. A 2 and B 2 act along 
the sides BC and AC, while A L \\AC and £ x || BC. As ^ and B t 
both pass through ?-, their resultant will pass through f . 





In consequence, the resultant of A 2 and B 2 , which passes 
through C, must pass through y, since C is the resultant of A lt B x 
and A 2 , B 2 . Now C^ is half the diagonal in the paraUelogram 
having AB as the other diagonal. In order that the resultant 
of A 2 and B 2 lies in the diagonal Cy the proportion 

A 9 5, 



A B 

must hold. — and -7-, however, are the specific stresses acting 

along the sections BC and AC. Hence the theorem: 

i/ the specific stresses acting on two different sections at a point 
are each resolved into components parallel to these sections, then 
the components acting along these sections are equal. 



226 



PROJECTIVE GEOMETRY. 



If the sections AC and BC are perpendicular to each other, 
A 1 and B ly A 2 and B 2 represent the normal and transversal com- 
ponents of the stresses. Hence the corollary: 

The transversal stresses acting on two perpendicular sections 
are equal. 

Taking at C any section CB and the stress A acting on it and 
drawing another section CA parallel to A, we have A 1 =A i B l =B, 
A 2 = B 2 = o. This gives the corollary : 

// the force A acting on a section CB is known, then the force B, 
acting on a section CA parallel to A , is parallel to CB. 




Fig. 97. 

2. According to this corollary for every section through a 
point C there exists a certain stress acting on this section, such 
that if the direction of this stress is considered as a section, the 
stress belonging to this section acts along the original section. 
This, however, is a clear expression for the involutoric character 
of the directions of sections and corresponding stresses. 

To prove this directly, keep in Fig. 97 the stress A and the 



APPLICATIONS IN MECHANICS. 227 

direction of the stress B || CB constant, and let the section BA 
turn about the fixed point B. The extremity A of BA traces 
on the section CA a point-range AA X A 2 . . . , so that the corre- 
sponding stresses B, B X1 B 2 . . . are proportional to the distances 
CA, CA X , CA 2 , .... In the force-polygon the extremities of 
the 5-stresses are marked by LL X L 2 . . . , and the corresponding 
C-stresses are ML, ML X , ML 2 , .... Now the distances NL, NL U 
NL 2 , . . . are proportional to the distances CA, CA X , CA 2 , . . . ; 
hence the projectivity of the pencils 

(B-AA X A 2 . . . )7\{M-LL 1 L 2 . . . ). 

Moving these pencils parallel to themselves so that M coin- 
cides with B, we have at B an involutoric pencil of sections and 
directions of corresponding stresses. 



§ 56. Discussion of this Involution. 

In Fig. 97 the sections CA and CB are both acted upon by 
compressions; in consequence the stress acting on the section is 
a compression. From the figure it appears clearly that corre- 
sponding rays of the involution in this figure move in the same 
direction. Hence, according to § 3, the involution is elliptic. 
The same is true if there are only tensions. In these cases there 
are no double-rays, i.e., there are no sections where there are only 
shearing {transversal) stresses. In all sections the material is 
either entirely under the influence of compressions or Under the 
influence of tensions. 

As every involution in .a pencil admits of two rectangular rays, 
it follows that through every point of a plane of stresses there are 
two sections on which only normal stresses are acting. In case of 
elliptic involutions these normal stresses are either both compres- 
sions or both tensions. 

If in Fig. 97 one section, say CA, is acted upon by a tension 
and the other, CB, by a compression, it is apparent that corre- 
sponding rays of the involution (§ 3) move in opposite directions. 



228 



PROJECTIVE GEOMETRY. 



The involution is hyperbolic and has two real double-rays 
(sections) in which only shearing stresses are acting. 

Considering two corresponding rays BA and C, they are al- 
ways separated by one of the double-rays, say d lt Figs. 98 and 99. 
If a compression acts on BA , it will be so when BA approaches 





Fig. 98. 



Fig. 99. 



d v But as soon as these corresponding rays have crossed the 
double-ray d lf the section A B is acted upon by tension. From 
this it follows that the material included by one angle formed by 
the double-rays is subject to tension only, while the supplementary 
part is subject to compression only. 

As the angles formed by the double-rays are bisected by 
the rectangular pair, it follows that the stress acting on one section 
where there is only a normal stress is a compressive force, while 
the stress acting on the perpendicular section is a tensile force. 

For a further discussion of these involutions and their exten- 
sion to space we refer to Ritter's Graphische Statik, Vol. I, pp. 
1-46, published by Meyer & Zeller, Zurich. 



APPLICATIONS IN MECHANICS. 



229 



§ 57. The Stress Ellipse. 1 Metric Properties of the 
Involution of Stresses. 

1. According to the previous section the specific stress acting 
on every section through a fixed point can be constructed as soon 
as the specific stresses acting on any two sections are known. 
In Fig. 100 assume these two sections, CA and CB, parallel to the 
x- and ^-axis of a Cartesian system, and let the variable section 
AB include an angle a with the positive part of the x-axis. As 
in Fig. 96, resolve the stresses A and B into transversal and 




Fig. 



normal components A ? , B 2 and A ly B v Designating the specific 
stresses determined by these components by t a , t b and n a , n b , we 
have A 1 = a-n a , B 1 = b-n b , A 2 = a-t a , B 2 = b-t b . Evidently t a = t b = /, 
say. The resultant C of A and B can also be resolved into trans- 
versal and normal components T and N with the corresponding 
specific stresses r and y, so that T = cr, N = cv. Let X and Y 
be the components of C parallel to the coordinate axes. Desig- 



1 Elegant graphic constructions for stress-ellipses may be found in Ritter's 
Graphische Statik, loc. cit. For a thorough analytic discussion see M. Levy's 
Statique Graphique, Vol. I, pp. 527-548 (Note IV). 



2 SO PROJECTIVE GEOMETRY. 

nating the specific stresses by $ and f), X = c£, Y = ct). From the 
figure we have 

a = c sin a, 6 = — c cos a, 
X=A 1 + B 2 = an a +b-t b , or 
X = c(n a sin a — t b cos a) and 

£=n a sin a— t b cos a. 
Y = B 1 -\-A 2 = bn b +at a , or 
F = — c(w 6 cos a— t a sin a) and 

7) = — n b cos a + t a sin a. 



Now 

v = — 7? cos a+6 sin a, hence 

v = w b cos 2 a— t a sin a cos a + w a sin 2 a— t b sin a: cos <x f or 

v = i(^ a +w 6 ) + icos 2a-(w 6 -w a )-isin 2a-(t b +t a ). 

Similarly, 

t = £ cos a + i? sin a, or 

r=w a sin a cos a— t b cos 2 a — n b sin a cosa-M a sin 2 a, or 

*=iK-fO sin 2a + i(/ a -/ )-|(/ a +/ 6 ) cos 2a. 

As t a =t b = t, we have finally 

(i) y= =i( W a+ W &)~£( W o"~ W &) C0S 2< * _/ Sm 20L > 

(2) T = i( W a _ W fe) Sm 2a— * C0S 2a * 

Designating the angle which the direction of C makes with 
the positive #-axis by /?, we have 

Q t) —n b cos a + / sin a 

tan p = -z= : : •, 

r £ n n sm a— L cos a 



or 



. / tan a — n. 

tan£=— \. 

r n„ tan a— t 



APPLICATIONS IN MECHANICS. 23 

Solving for tan a, we get 

(3) tan a = 



t tan /?— n b 



n a tan@—t' 

which clearly shows the involutoric character between the direc- 
tions of a section and the. stress acting on this section. This is 
in agreement with the geometric discussion of stresses in § 55. 
For the double-elements of this involution we have 



n jn 2 — 2tm-\-n b =0, 



/ x t±Vt 2 —nn h 
(4) w=— ^~ b . 

These values for m are real when n a and n b have different 

signs; in this case the involution is hyperbolic. If n a and n b 

both have the same sign, and n a n b >t 2 , then the involution is 

elliptic. For n a n b =t 2 the involution is parabolic and tan a = 

t 
— = const.; i.e., the stresses all act in the same constant direction. 

This is the case in a rod under tension or compression exclu- 
sively. 

2. Letting v x and v 2 be, the normal specific stresses on two 

perpendicular sections determined by the angles a and a——, 

and ± r a the transversal specific stresses in these sections, from 
the formulas for v and r we get 

T « 2 — v i v 2 = * 2 — n a ti b = const , 

To get the rectangular pair of the involution, we form 

/ tan a—n. 
tana \ 

tan («-#= f: . 

1 + tan a 



n a tsLna—t 



232 PROJECTIVE GEOMETRY. 

1 • A n T ^ tai1 a ~ n h 

In this a—p =— t if i + tan a- — = o, or 

' 2 n a t&na — t 

t tan 2 a + (n a — n b ) tan a — t = o, or 

tana = ^~^ ±V( ^"^ )2+4 ^ 
2* 

an expression which is always real. 

From this, and also from the expression for r = o, follows 

(5) tan 2«= . 



3. We shall now find the locus of the extremities of the specific 

c 

stresses — acting on all sections. Its coordinates are evidently 

£ and tj when referred to the point of application as an origin. 
From 

(6) n a sin a — t b cos a = £, 

(7) / a sin a— n b cos a = y, 
the expressions for sin a and cos a result : 



sin a = 


"»«»& 


-t 2 ' 


cos a = 


/£- 

n a n b 


-z 2 ' 



and since sin 2 a + cos 2 a = 1, the required equation 

(8) £ 2 K 2 +^ 2 )-2^K+^)^+^K 2 +^)-(V6-0 2 = 

results. In this equation 

(n a *+n(n b *+t>)-(n a +n b )H>=(n a n b -t>y>o-, 
it represents, therefore, an ellipse, the so-called stress-ellipse. 



APPLICATIONS IN MECHANICS. 233 

From analytical geometry, § 31, the angles d and 0+— of the 
axes of this ellipse are determined by 

(9) tan ^=V+^-(V+^) = K^' 

Hence, according to (5), we have the theorem: 
The axes of the stress-ellipse coincide with the rectangular 
pair of the involution of stresses around the center of the ellipse. 
From 

v = \{ n a + n b)~ i( n a ~ n b) cos 2a— /sin 2a 

we find, by differentiation with respect to a, the condition for 
maximal and minimal normal stresses, 

/ sin 2a(n b —n a )—2t cos 2a =0, 
or 

2t 

(10) tan 2a = . 

n a - n b 

Hence, according to (5), the theorem: 

The maximal and minimal normal stresses occur on the sections 
of the rectangular pair of the involution, or on the axes of the stress- 
ellipse. In these sections r = o. 

In a similar manner, from 

r== i( n a~ n b) s i n 20i ~l cos 2a 

we find for the maximal and minimal shearing stresses the condi- 
tion 

n— n u 



(11) tan 2a 

hence, from (10) and (11). 



/ _ 



2/ 



/ 1Z , K 

tan 2Ct-tan 2a — — 1. 2a = 2a,±—, a =a±— , or: 

2 4' 

The directions of the maximal and minimal shearing stresses 
bisect the angles formed by the maximal and minimal normal 
stresses and are equal {except as to sign). 



234 PROJECTIVE GEOMETRY. 

4. The directions ft and ft of the stresses corresponding to 
two rectangular sections with the inclinations a and a+- are, 
according to (3), determined by 



/ n n t tan a—n, 

(12) tan ft = 



tan a—V 

u 

t rt 

(13) tan ft = 



/ cot a + n b 



n a cot a-\-t 

_ , x 2 tan ft— w, w tan ft— 2 

From( I2)) ta*--^-^, cota =-^. 

Substituting this in (13) and reducing, we get 
(ta\ t»n ft - <(». + »>) tan fl-(V+ P) 



From this formula follows at once : 

r&e directions of pairs of stresses corresponding to pairs of 
perpendicular sections form an involution. 

This involution is identical with the involution of conjugate 
diameters of the stress-ellipse. 

To prove the second part of this theorem, form the equa- 
tion of the polar 

(i5)^i(V+0-(^+^i)K+^+^i(V+ /2 )-K^- /2 ) 2 =° 

for any point (£„ ^), for which 7^ = tan ft, with respect to the 

stress-ellipse. 

For the point infinitely distant in the direction of ft there 

still is ^ = tanft and £ x = 0°, ^= 00. Hence 

£K 2 +' 2 )~ 0?+ £ tan ft)(n a +» 6 ) + t? tan ft(» 2 +/ 2 ) =0, 



APPLICATIONS IN MECHANICS. 235 

or 

■H t(n a + n b )tznp-(n b *+t*) 



£ (n a 2 +t 2 ) tanA-/(» a +n 6 ) ' 



which is nothing else than tan /? 2 in (14), q.e.d. 

5. The normal stresses, v u v 2 , on two perpendicular sections, 

determined by the angles a and a——, are 

»l. = h( n a+ n b)- h{ n a- n b) C0S 20i ~ t Sm 2(X > 

v 2 = i(n a +n b ) + %(n a -n b ) cos 20. + 1 sin 2a. 
Adding, we get 

(16) v 1 +v 2 = ^ a +w 6 = const. 

Hence the theorem: 

The sum 0] the normal stresses acting on two perpendicular 
sections is constant and equal to the sum oj the maximal and mini- 
mal normal stresses. 

6. Between the angles ft and a which the directions of a sec- 
tion and the corresponding stress make with the positive x-axis 
the involutoric relation 

„ s n t tan a— n, 

(17) tan£=-— \ 

exists. The central ray of the involution, for which /?=o, is deter- 

n 
mined by tan a = ~. Designate this value of a by y, so that 

n 

-~ = tsm y. Take a line parallel to the ray for which /?=o, at a 

distance p from it, and project the involution of stresses on this 
line. Then, from Fig. 101, A CBC = const. To determine this 
constant, we have BC = BD—CD, AC=CD+DA; hence 



AC-BC = 



tan /? tan y) \ tan y tan 



a) 9 



236 PROJECTIVE GEOMETRY. 

or, after reducing, 



(18) 



AC-BC = -p 2 a \ =k, say. 



In a similar manner, for the constant of the involution of 
conjugate diameters of the stress-ellipse we find 



(19) 



(fl 71 t^} ^ 

A A • 5 A = - P 2 \;/ +f y = K say. 



Assuming that n a and n h are normal stresses on two perpen- 
dicular sections, then t = o. Without loss of generality we may 




also assume p = i, so that (18) and (19) become 



n in 

(20) k= °, &!=-(- 



where n a and n b now designate the maximal and minimal normal 
stresses. Hence, when the stress-ellipse is known, it is not diffi- 
cult to construct the involution of stresses. 



§ 58. Examples. 
1. A plane linear deformation is defined by the equations 



<i) 



f x f = ax J r by y 
I y' = a x x+b x y. 



APPLICATIONS IN MECHANICS. 237 

Referring the points (x, y) and (V, /) to an oblique system 
of coordinates (£, rj) having the same origin and whose axes 
include the angles a and a x respectively, we have 



ji = ccosa 
( v = c sin a 



rj cos a 1} 

-{-rj sin a v 



Applying this to the points (x, y) and (V, /), we get, accord- 
ing to (1), 

£' cos a+rf cos a 1 = $(a cos a + & sin a) + r)(a cos 0^+6 sin ctj, 

£' sin a: +7/ sina: 1 = f(a 1 cos a-f-^ sin a) + 77 (a x cosa 1 +6 1 sina^), 

or, solving for c' and 7/, 

{ £(a cos a sin a t + 6 sin a: • sin a t — a x cos a cos a t 



sin (a ± —a) 

— & x sin a: cos aj + t?[& sin 2 a^— a x cos 2 a t 
+ (a— 6J sin a^ cos aj } , 

19' = -; — 7 c { £[b sin 2 a t — a, cos 2 a+ (a— 6.) sin a cos al 

' sin (a— ctiV v 1/ j 

+ 9 (a cos «! sin a + b sin a x sin a— a x cos a x cos a 

— b 1 sin a x cos a) J . 

In these expressions we can choose the angles a and a x in 
such a manner that tan a and tan c^ are the roots of 

6-tan 2 a+ (a— 6 X ) tana— ^ = 0, 
so that 

tan a = &-a+vU-a) 2 +W> 

2& 



26 



238 PROJECTIVE GEOMETRY. 

Under these conditions the coefficients of r; and $ in the ex- 
pressions for £' and 7/, respectively, vanish and the linear trans- 
formation in this system of oblique coordinates assumes the form 

[ g, __ - 2{ab- 2a x b) + (a 2 + b 2 ) + (a+ bjVfa- a) 2 + \a J 

, x J V(b 1 -a) 2 +4a 1 b 

I , = 2(fl6 1 -2g 1 6)-(g 2 +V)+ O + 6 1 ) \/(6 1 -g) 2 +4fl 1 6 > 
I V(\— a) 2 +4a x 6 

From these formulas follows that the linear deformation (1) 
may be considered as two consecutive stretches along two oblique 
axes or directions. The angle <j) formed by these axes is deter- 
mined by 

tan,£ = ^S±4^. 
b—a x 



2. Evidendy the rectangular transformation 

(4) 



x' = ax, 



is a special case of (3). 

The elongations along the x- and ^y-axes are 

x'—x y r —y 7 

=a—i and — = 0«— 1. 

x y 

We can also write (4) in the form 

f x ' = x-\- (a— i)x, 
(5) [y=y+(b 1 -i)y. 

Consider (5) as the analytical expression of a strain in a very 
thin plate which is assumed to have the property of a perfect 



APPLICATIONS IN MECHANICS. 239 

solid. Then a— 1 and b 1 —i are very small numbers. The 
strain-ellipse has the equation 

a 2 x 2 +b 2 y 2 = r 2 . 

By the linear deformation certain stresses are produced which 
according to Clebsch 1 may be expressed in terms of the strains 
a— 1 and b x — 1. In our special case there are no shearing stresses 
along the x- and y-axes, so that in the formulas 2 

N x = Xd+2iia, N 2 = Xd+2pb 

we have 6 = a J r b 1 — 2, X = fi for a perfect solid, a=<z— 1, b = b 1 —i i 

N 1 -n a , N 2 = n b ; hence 

n a = t(3a+b 1 -4), 

w 6 =^0+3&i-4), 
and the equation of the stress-ellipse 

x 2 y 2 



P( 3 a+b- 4 ) 2 < X\a+ 3 b- 4 y 



§ 59. The Rectangular Pair of the Involution of Stresses in 

Nature. 

In the sections corresponding to the rectangular pair of the 
involution only normal stresses are acting and these represent 
the maximal and minimal normal stresses. If at the point con- 
sidered we advance an infinitesimal amount in the direction of 
one of the conjugate rectangular sections, for instance that for 
which the normal stress is a maximum, we can at this place, 
infinitely close to the first, again construct or calculate the two 
rectangular pairs of the involution. On the section for which 

1 Theorie der Elasticitat fester K or per, Leipzig, 1862. 

2 Minchin: Treatise on Statics, Vol. II, p. 435, fourth edition. 



240 



PROJECTIVE GEOMETRY. 



the normal stress is a maximum we can again advance an infini- 
tesimal distance and then construct the two conjugate normal 
sections, etc. In this manner a curve is obtained along which 
the normal stresses have their maximal values. If these stresses 
are tensions which are greater than the elastic limit of the material, 
then the material will rupture along this curve (maximal tension 
curve). In a similar way a curve may be drawn through the 
point along which the normal stresses have their minimal values. 
If the involution is hyperbolic, this curve is a maximal compres- 
sion curve, since the stresses along this curve are maximal with 
reference to the compressive stresses. 




MAP 

OF THE 

ARAPAHOE GLACIER 

AUGUST 1902 

i v* MILE 1 A 



Fig. 102. 



This case of a hyperbolic involution is shown in Fig. 102, 
which shows the crevasses of Arapahoe glacier. 1 The stream- 



1 From a drawing by Professor N. M. Fenneman in an article: The Arapahoe 
Glacier in IQ02, Journal of Geology, Vol. X, p. 841. 



APPLICATIONS IN MECHANICS. 



241 



lines represent the curves of maximal compression, while the 
crevasses cutting the stream-lines orthogonally represent the 
maximal tension curves. The case of an elliptic involution 
where there are only tensions is illustrated by the cracks which 
form on a heavily varnished surface or in a mud-bed which is 
drying up. In this case only tensile normal stresses act on the 
rectangular pair. One is a maximum, the other a minimum. 
We should therefore expect that the maximal tension curves 
would form a system of more or less parallel curves. This, 
however, does not occur, as is seen from Fig. 103, in which the 




Fig. 103. 

cracks intersect or meet orthogonally. This can be explained 
in the following manner : After a crack has formed, the maximal 
stress and strain normal to the crack has been relieved, so that 
the former minimal normal tension along the crack now becomes 
the maximum. Hence the next rupture will be orthogonal to 
the first crack. 1 



1 See my article on this subject in the American Mathematical Monthly, Vol. 
VII, pp. 134, 135. Further examples may be found in Ritters Graphische Statik, 



Vol. I, pp. 128-134. 



242 PROJECTIVE GEOMETRY. 

Realization of Collineations by Linkages. 

§ 60. Introductory Remarks. 

We have seen that collineations may be produced analytically 
and synthetically by different methods. In what follows a num- 
ber of linkages will be described by which collineations may be 
realized kinematically. Linkages, like pantographs, translators, 
etc., devised for some special purpose have been known for a 
long time. The history of linkages in connection with the theory 
of geometrical transformations, however, dates back to the year 
1864, when Peaucellier found a rigorous solution for the prob- 
lem to describe a straight line by a link-motion. 1 Since that 
time a number of geometers, among whom the English Sylvester, 
Hart, Roberts, Cayley, and Kempe occupy the foremost places, 
have made a systematic study of linkages and their geometric 
properties and have found a great number of important results. 
Among these investigations probably the most interesting are 
those of Kempe and Koenigs. The first proved the theorem 
that it is always possible to find a linkage so that one of its points 
describes any given algebraic curve. Koenigs generalized this 
by proving that every algebraic surface and curve may be described 
by a linkage. As a result of these interesting theorems it is not 
difficult to prove that any algebraic transformation between any 
number of variables may be realized by linkages. 2 The diffi- 
culty lies in the actual construction of such linkages. Recently 
Koenigs has invented a linkage which realizes a general projec- 
tive transformation in a plane. 3 

1 Nouvelles Annales de Mathematiques, 2d series, Vol. Ill (1864), p. 144. 

2 Koenigs: Legons de Cinematique (1897), pp. 262-307. See also Transac- 
tions of the American Mathematical Society, Vol. Ill, pp. 493-498 (Oct. 1902), 
where the author proves that any number 0} algebraic relations between n complex 
variables may be realized by a plane linkage. 

3 Comptcs Rendus, Vol. CXXXI, p. 11 79. The different cases of collineation 
were worked out by Hermann Emch in his Master's Thesis at the University of 
Colorado, 1902. 



APPLICATIONS IN MECHANICS. 243 

To have a definite idea about the character of the plane 
linkages to be considered I set down Koenigs's definition: 

A plane linkage {systeme articule, Gelenkwerk) is a combina- 
tion of plates or plane figures subject to remain in one and the 
same plane, among which a certain number are connected to each 
other by hinges or pivots perpendicular to the common plane. 

In this definition it is assumed that the links move by each 
other without interference, which means that the links, consid- 
ered as material, he in a series of close parallel planes. 

Even- linkage is constructed in such a manner that one of 
its pivots is fixed and represents the origin O, while others repre- 
sent the algebraically related variables. The points of the link- 
ages will always be designated by the same letters as the corre- 
sponding variables. 

Two or more linkages each involving two variables may be 
combined in the following manner: Suppose Z, L lf L v . . . , L n 
are linkages realizing the transformations 

Let the origins of all these linkages coincide; attach the pivot 
u n of L n to the pivot u n of L n _ x \ attach the pivot U n _ 1 of L n _ 1 to 
u n _ 1 of L n _ 2 , and so forth; finally the pivot u x of L x to U t of L. 
Then the point u of L evidently realizes the compound trans- 
formation 

«=/{/.%-.. /»-i(/»(z))]} =F(i). 

Linkages involving more than two variables may be similarly 
combined. 

The range of effectiveness of a linkage is, of course, limited 
to a certain finite portion of the plane. This range, although 
in some cases small, alwavs exists. 



244 PROJECTIVE GEOMETRY. 



§ 61. Analytic Formulation of the Problem. 

We shall consider only the most important cases of collinea- 
tion. A great number of special cases will be left as exercises 
for the student. 

The most important cases are the linear trans j or mation and 
perspective. A linear transformation 

f x 1 = ax+by+c i 
^ [y^dx+ey + f 

may be considered as made up of four subgroups: (i) the- two- 
termed translation, (2) the one-termed rotation, (3) the two- 
termed dilation, 1 (4) the one-termed elation. By a translation 
(p, q) and a rotation (</>) the point (x, y) is changed into (V, /). 



(2) 

A dilation 

(3) 



x f =x cos <f>— y sin </>+^>, 
y' =x sin <j> +y cos <f>+q. 



x" =ax f , 

y"=W 



changes (V, /) into (#", /') : 

x" =ax cos <j)—ay sin cf)-\-apj 

y"=P- 

Finally by the elation 



^ I y"=[$x sin <£+/fy cos </> + /?<?. 



x. =x n ' + Ay", 

(5) i „. _ i// 



we get 

J x x = (a cos $-M/? sin <j>)x-\-(^ cos <£— a sin (f))y-\-ap+A[$q, 

\yi = P s i n 4>- x +P cos <j>-y+l3q, 

1 Term used by S. Lie, loc. cit. It is equivalent with dilatation, p. 60. 



APPLICATIONS IN MECHANICS. 



245 



which by properly choosing a, /?, X, <j>, p, q, the six parameters, 
may represent any linear transformation of (x, y) into (x lt y x ). 
To prove this let 

a cos <j>+Xfi sin <f> = a, 
Xft cos <j)—a sin <f> = b, 

ap+A[3q = c, 
/? sin = d, 

/? cos <£ = e, 

k=n 

which represent six equations with six unknown quantities a, /?, 
^> 4>) Pi 9.- Solving, we get 



ae — bd „ , 



ad-\-be 



d c(d 2 +e 2 )-}(ad+be) 
^ = arctan-, j = flg _ w , <7 



Vtf 2 + 



Substituting these values in (2), (3), (5), we obtain a trans- 
lation with rotation, a dilation, and an elation which in succession 
transform (x, y) into (V, /), (V, /) into (x", y"), and finally 
(x", y") into (X, v x ) in such a manner that (x l7 y x ) is connected 
to (x } y) by the linear transformation 



(7) 



x 1 = ax+by+c, 
y 1 = dx+ey+f. 



Applying to (x lt v x ) the perspective 
(8) 



x' = 



y= 



^1+^1+ A' 



24b PROJECTIVE GEOMETRY. 

we have 



; ax+by+c 

x == 



(9) i 



(d x a+ e x d)x+ (dj)+ e x e)y+j x ' 

dx+ey-\- } 
(d x a+ e x d)x+ (d x b+ e x e)y+) x y 



which may represent any projective transformation. The linear 
transformation is a six-termed and the perspective a three- 
termed group, so that their combination (9) contains nine param- 
eters, although the general projective group is eight-termed. 
This is due to the fact that both the linear transformation and 
the perspective contain the one-termed group of similitudes as a 
subgroup. These considerations, which may be found in a little 
different form in § 19, have been repeated here for a clearer under- 
standing of what follows. 

We shall now proceed to describe linkages realizing the trans- 
formations in question. Theoretically only such linkages should 
be admissible in which a link joins two and only two points. 
In other words, no three points in a straight line should be ad- 
mitted a priori. It is, however, very useful for practical pur- 
poses to make this last assumption. For some transformations 
we shall construct more than one linkage in order to show the 
advantage which one or the other may have. 

Combining the linkages involved in the linear transforma- 
tion and in perspective according to the scheme explained in 
the last part of § 60, a compound linkage for a general collinea- 
tion is obtained. 

§ 62. Peaucellier's Inversor. 

In our particular investigation of link-motions the problem 
to draw a straight line theoretically correct is of the greatest im- 
portance. This can be done by Peaucellier's inversor (loc. cit.) 
or by Hart's linkage (Koenigs's Cinematique, p. 267). Peau- 
cellier's inversor is of greater principal value and will be described 
here. 



APPLICATIONS IN MECHANICS. 



247 



It consists in the first place of a rhombus ABPP' and two 
equal links AO and BO. In all these points the links are joined 




T --' 



L 



Fig. 104. 



by pivots, Fig. 104. Designating OA = OB by a, AP = PB = BP r = 
P f A = b, A A' = c, we have OA' = \(OP+OP'),A'P = i(OP'- OP); 
i(0P+0P') 2 = a 2 -c 2 ; c 2 = b 2 -\(OP'-OP) 2 , or l(OP+OP') 2 - 
HOP'-OPy^a 2 -!) 2 , or finally 

OP-OP'=a 2 -b 2 . 

Hence, if O is kept fixed, the points P and P r are inverse 
with respect to a circle having O as a center and V a 2 — b 2 as a 
radius. If now P describes a circle with M as a center and OM = r 
as a radius, we have OP-OP' = OT-OQ, or OP/OT = OQ/OP'; 
consequently aOPTcoaOQP'; and since Z OPT = 90°, also 
Z OQP' will be a right angle. Hence, when P describes said 
circle, P ; will describe a straight line perpendicular to the direction 
ofOM. 

For the limiting position OSRS' of the inversor we have 



or, since 



QS'=VOS' 2 -OQ : 



2 J\2 

OS' 2 =(a+b) 2 and 0(2 = °——, 



QS' = —\/{a+ b) 2 4 r 2 - (a 2 - b 2 ) : 



2 4 8 



PROJECTIVE GEOMETRY. 



Of course the lengths a, b, r must be chosen in such a manner 

a—b 
that the linkage is movable. The conditions are r> , fol- 
lowing from (a+by^r 2 — (a 2 — 6 2 ) 2 >o, and, for the case that the 
straight line shall not cut the circle of inversion r< JV 'a 2 — b 2 . 

Ex. i. Show that when M, P, and A are in a straight line, 
AP'LS'Q. 

Ex. 2. If the whole linkage is in a vertical plane and OM 
vertical, the linkage remains in equilibrium under the action of 
any weight suspended at P' '. 

Ex. 3. If a and b are given, what value must r have to make 
QS' a maximum ? 



§ 63. Pantographs. 

1. Inversor Pantograph. 

By means of Peaucellier's cell ABPQO, a part of the inversor, 




Fig. 105. 



we can locate for every point P a point Q, so that OP • OQ = a 2 — b 2 . 
Applying another cell, A x B&'P'O x , for which O&' • OJ>' = a 2 - V, 



APPLICATIONS IN MECHANICS. 



249 



and letting 1 coincide with O, and Q with Q', Fig. 105, then 
0&' = OQ\ and by division 

OP a 2 -b 2 



OP' 



V 



Hence, when P describes a figure, P x will describe a similar 

figure. Choosing O as the origin of Cartesian coordinates and 

a 2 -b 2 
designating the constant ratio — 2 — j- a by /c, then when O is fixed, 

the linkage of Fig. 105 will realize the transformation of P(x, y) 
into P'(x f , /): 

x f = kx, y = Ky. 

2. Sylvester , s Pantograph. 1 

Take any two similar triangles, Fig. 106, OAA' and APB 
pivoted at A , with AA'OA= z BAP and /.A'AO = Z £P4 . 




Now Z(L4'P'=Z4'.45, zei , 0=Zi , 04+Zi'iO; hence 
/.QA'P'+ £QA'0= £A'AB+ ABAP+ /.A'AO, or ZOzl'P'^ 
Z0^4P. But there is also 

OA AP AP 
OA'~ AB~ A f P n 

1 Nature, 1875, p. 168. 



250 PROJECTIVE GEOMETRY. 

OA OP 

hence AOi'P'^AOi?. From this Qj, = Qp, and ZAOP = 

LA'OP', hence also £POP'= IAOA' and &POP'co aAOA'. 

Consequently when P describes a figure and O remains fixed, 

P f will describe a similar figure. The ratio of similitude between 

the figures traced by P and P' is OA/OA'. Turning the figure 

traced by P f negatively through an angle = AA'OA it will come 

similarly situated with the figure traced by P with respect to 

the center O. Designating by cf> the angle A'OA, and by p the 

OA ' 
ratio TyTi Sylvester's pantograph will realize the combined groups 

of rotation and similitude between P(x, y) and P f (x', /): ' 

x' = p(x cos <f>— y sin 0), 

y = p(x sin <j)-\-y cos <f>). 

This becomes a pure rotation when p= 1 ; i.e., 0^4 =0^4'. 

The arrangement of this linkage is a little different from 
Sylvester's original pantograph, but does not essentially differ 
from it. 

3. P. Scheiner's Pantograph (1631). 

Schemer's or the ordinary pantograph appears in the market 
under many different forms. One of the simplest is illustrated 
in. Fig. 107. Two equal sets of links PQ, PR and CP', CO 



pivoted at P and C are placed in such a manner that P is in a 
straight line with O and P', andP()||CP', PR\\CO. In this 
position pivots are also placed where PR and CP', and PQ and 
CO meet. From the figure it appears at once that when O is 
fixed and P describes a figure, then P' will describe a figure 



APPLICATIONS IN MECHANICS. 251 

similar to and similarly situated with the first. The linear ratio 
between the two figures is OP /OP'. To make different values 
for this ratio possible the links may be divided into equal parts, 
as shown in Fig. 107. Wishing to enlarge a figure in the linear 
ratio 4 : 3, set the pivots where PR and CP', and PQ and CO meet 
at the marks 6, so that OP'/OP = S : 6 = 4 : 3. In a similar manner 
arrangements for any other ratio may be made by properly 
dividing the links. 

Although Schemer's pantograph is the simplest of all panto- 
graphs and consequently exclusively used for practical purposes, 
it has the theoretical disadvantage of not being a pure linkage. 
Indeed, in Fig. 107 it is assumed that a link joins three given 
points in a straight line. The first two pantographs described 
are pure linkages. 

§ 64. Rotator and its Combinations. 1 

1. To realize a rotation through an angle of a point P(x, y) 
into P'(x', /), so that 

x' = x cos 4>—y sin <j>, 
y'=x sin <j>-\-y cos <f>, 

Sylvester's pantograph in the case p = i may be used. Another 

linkage for the same purpose, Fig. 108, is obtained by taking 

two isosceles triangles OAC and OBD 

pivoted at O, the coordinate-origin, with 

/AOC = /BOD = <j> and'AO = CO=BO = 

DO. Attaching the links PB = PC, P/A = 

P'D, pivoted at P and P' respectively, and 

all equal to AO, two equal rhombs OBPC 

and OA P'D are obtained. Hence /AOP = 

I BOP', /POB = ZP'OA, and /BOP' + 

APOB=ZAOP+ /.COP, or 

/POP'=/AOC = cf>. 

1 A paper on this linkage and its applications was presented to the Am. Math. 
Soc. in Chicago, Sept. 1902, and was published in Vol. I of The University of 
Colorado Studies, April 1903. See also Transactions of the Am. Math. Soc., 
Vol. IIT, No. 4, pp. 493-498, Oct. 1902. 




252 PROJECTIVE GEOMETRY. 

Furthermore, P'0=PO. The linkage of Fig. 108 can therefore 
be used to perform the proposed rotation. 

2. The foregoing linkage may be used for various purposes. 
In the first place when O is not fixed, we have in the three pivots 
P, O, P f a variable isosceles triangle which in all its deforma- 
tions remains similar to some original size. When O is fixed and 
P describes a straight line or a circle, P r also describes a straight 
line or a circle respectively. Making (f> = go°, and taking two 
equal rotators with the points P and P r attached, a variable 
square is obtained. 



§ 65. Translators. 

One of the simplest devices for translation is that of Kempe. 1 
It consists of three parallel equal links AD, BC, PP' which are 

connected by CD = \\ A B and CP' = 

\\BP. Letting A coincide with the 
coordinate-origin and designating the 
coordinates of D by a and b, then for 
the coordinates (x, y) of P and (a/, /) 
of P' we have from Fig. 109 
x' = x+a, 
y'=y+b. 
A translator which is more general 
is obtained from a linkage which was originally invented to per- 
form the addition of any complex variables. 2 It consists of 12 
links,. Fig. no, of which OF = \\CE = \\BD; FP' = \\EG = \\DP; 
OC = \\FE = \\P r G\ CB = \\ED = \\GP. It is evident that OBPP' 
will always be a parallelogram no matter how the linkage may 
be deformed. Hence, keeping B and O fixed, P' represents in 
every position of the linkage a translation of P equal to BO and 
in the direction of BO. 




1 How to Draw a Straight Line. 

2 Transactions, loc. cit. 



APPLICATIONS IN MECHANICS. 



253 



§ 66. Linear Transformation. 

1. By a combination of rotator and translator it is possible 
to realize a general motion in the plane. According to § 61 




Fig. i 10. 

the next group in making up a linear transformation is the dila- 
tion 

x" =ax f . 

?'-&■ 

The linkage for this transformation is shown in Fig. in. 1 
Let OA=x / . By a Schemer pantograph (which we choose for 
the sake of simplicity) , in which OB/OA = a and consisting of 
the links OC, CB, DA and AE, a point B is realized for which 
BO=x" =ax f . One of the points A and B is kept on the x-axis 
by a Peaucellier inversor. To A and B attach a translator 
ABIP'VLMNH. Produce the link NB arbitrarily to R and 



1 This is essentially the arrangement of Hermann Emch in his thesis, loc. cit. 



254 



PROJECTIVE GEOMETRY. 



complete the rhombus BRST. To BR and BT attach at B the 
equal right-angled triangles RBF and TBG, so that BF = BG. 
Complete the rhombus BFGP" '. From the figure follows easily 
that IFBG = IRBT and that P"BLBS, LOB. 

Now use FB and FP" as links of a second Scheiner panto- 
graph, and attach the links IK and IU in such a manner that 




Fig. hi. 



BF\IK = P"F\ITJ = BP" \IP", and P"B/BI=p. The point / 
is now collinear with P" and B, and as P'A\\IB it follows that 
P'ALOA. Making P'A=y f , we find IB = P'A=y'; P"B = 
fi-IB=py. The coordinates of P" are therefore 



BO=x"=ax', 
P"B=y"=Py', 



and we have constructed a linkage realizing dilation. 

2. The last group to be considered is the one-termed elation. 
Take two rhombs AFP n F and ACBD with the common joint or 
pivot A ; join E and C, and i? and Z) by two equal links EC = FD 9 



APPLICATIONS IN MECHANICS. 



255 



so that /.EAC= Z.FAD = — . From plane geometry there fol- 
lows easily LEAF = LCAD; i.e., that the two rhombs are similar; 
further, that F"A _L AB, no matter how the linkage may be 
distorted. This linkage realizes, therefore, a variable right tri- 
angle F" AB whose angles are constant. Joining in a symmetrical 
manner the rhomb BHP 1 G=AEP ff F to the previous linkage 
(CG = CE, HD = FD), a variable rectangle ABF"F X is obtained 
whose sides have a constant ratio, Fig. 112. 




0- 



Fig. 112. 



This linkage may be used to solve mechanically two interest- 
ing cases of collineations in a plane. If by two Peaucellier 
Inversors the points A and B are forced to describe the same 



256 PROJECTIVE GEOMETRY. 

straight line s, in which an arbitrary point is taken as the origin 
of a Cartesian coordinate system, and 5 itself is assumed as the 

AB 

#-axis, we have, since ■ . p ,, =ni (constant), for the coordinates 

x lt y x of P- 1 in terms of those of P" \x n ', /'): 

x l = x" ' + my n ', 

y.=y", 

which represents the required elation. 

If the rhombus BGHP 1 and the links CG and DH are attached 
below s, so that the point P x will fall on P 2 , then the coordinates 
of P 2 are 

x z =x f, -\-my fr , 

y»— /', 

which represents oblique axial symmetry. Combining the link- 
ages for rotation, translation, dilation, and elation as explained 
in § 60, a linkage for the linear transformation is obtained. 

Ex. 1. Construct a linkage for the transformation (oblique 
axial symmetry): 

x r = x+my } 

? = -y. 

Ex. 2. Construct a linkage for the transformation (orthogonal 
axial symmetry) : 

x'=x, 
y' = -y. 
Ex. 3. Construct a linkage for the special dilation: 
x' =ax, 

y= y . 



APPLICATIONS IN MECHANICS. 257 

Ex. 4. Construct a linkage for the central symmetry: 

Ex. 5. Draw the combined linkage for a general linear trans- 
formation. 

Ex. 6. Determine the ranges, or limits of the areas, covered 
by Sylvester's pantograph, the rotator, the translators, and 
Scheiner's pantograph as used in the linkage for dilation. 

§ 67. Perspective. 

1. Mechanisms by which the perspective of any plane figure 
may be drawn are known in various forms. One that is in prac- 
tical use is the " perspectivograph " invented by H. Ritter. 1 In 
this mechanism pivots are kept on given straight lines by grooves 
so that it combines link- and sliding-motions. Another "per- 
spectivograph" in which two ellipses are used and which also 
combines link- and sliding-motions was described by the author 
some years ago. 2 Probably the most important linkage-realizing 
perspective has been invented by Koenigs, 3 and, as it does not 
use slide-motion, will be described here. We must, however, first 
describe Kempe's reversor which Koenigs uses as an auxiliary 
linkage. 

2. Kempe's Reversor. 

In Fig. 113 consider the linkage OBCD in which OB and CD 
are equal and cross each other, and also OD = BC. Designating 
the variable point of intersection of OB and CD by X, this linkage, 
which is called counter-parallelogram, has the property that for 
any deformation a BOD= a DCB; aOXD=aCXB. On DC 
choose a pivot E in such a manner that DE:DO=DO:DC, so 
that the triangles OCD and EOD are similar. Then with OD 

1 See Geometrische Trans formationen by Dr. K. Doehlemann, pp. 199-204, 
Leipzig, 1902. 

2 The Industrialist, Vol. XXV, pp. 237-240, Manhattan, Kansas, 1899. 

3 Comptes Rendus, Vol. CXXXI, p. 1179. 



2 5 8 



PROJECTIVE GEOMETRY. 



and ED as given links complete the counter-parallelogram ODEF, 
in which FE = DO, FO=DE. Thus aEOD=aOEF and 
similar to &OCD= &COB. Hence, in every deformation, the 




Fig. 113. 

counter-parallelograms OBCD and ODEF are similar, and as 
a BOD = A DCB, it also follows that a DOF = A FED. By means 
of this reversor it is possible to keep two links BO and FO at 
equal angles from a given link DO. By a similar construction 
two other links, B'O and i^'O, symmetrical to DO may be attached, 
and it is clear that their motion is otherwise independent of that 
of BO and FO. We have therefore a linkage in which in every 
deformation 

Z.BOB f =AFOF ; . 

Kempe's reversor may be extended to realize any number of 

equal angles, lBOD= lDOF= AFOH= For the details 

of this we refer to Koenigs's Kinematics, loc. cit. 



Koenigs' Perspectivograph. 



3. Introducing polar coordinates, x = rco$ 6, v = rsin 6, in the 
formulas for a perspective transformation 



(1) 



y 



dx+ey+f' 

y 

dx+ey+F 



APPLICATIONS IN MECHANICS. 



259 



we get, since 0' = = arc tan I - J =arc tan f -^ I, 



(2) 

and 

(3) 



7-(dcos#+esin#) + /' 
- = d cos #+e sin 0+-. 

f Y 



(4) 



Putting d=-- sin 



tan -1 )" ). (3) becomes 
r f r 



— cos </>, so that a 



Vd 2 + 



and 



sin (0+4>). 



Now take two Peaucellier inversors OABQP and OCDQ'F', 
Fig. 114, and by means of Kempe's reversor as described in 




Fig. 114. 



Pig. 113 keep ZAOC = I BOD. This can be done by properly 
choosing B', F', B, F of Fig. 113 on BO, AO, DO, CO of Fig. 114, 



2 0o PROJECTIVE GEOMETRY. 

respectively. Let OP = r, OP f = r f ; p and // the squares of the 
radii of the circles of inversion of the two inversors; OQ = p, 

OQ' = p'. Now OPOQ = p, or r-p = p, hence— =— ; 

OP r • OQ' = //, or r'-p' = //, hence — = — 7 . So far, no particular 
values are assigned to p and //, so that we can choose /* = /•//; 
— =— . Equation (4) now becomes 

(5) ^~^ =_J J sin(^+^). 

To the two inversors attach a Kempe translator OO'EFRQ, 
where O' is on the v-axis, and join R to Q' by the link RQ' =RQ. 
Let /.POX = 6+<f>; then lQRQ f = 2(6+ <f>), and QQ' = ff- p = 

2 RQ- sin (d+4>). Hence, taking 00' = RQ = -- ^- = ^-Vd 2 +e 2 , 

2 (JL 2 

the points P' and P realize the proposed perspective transforma- 
sion, since the linkage satisfies all conditions of equations (1) or 
(2), or their equivalent (5). 

Ex. 1. Modify the linkage so that P and P' describe similar 
figures. 

Ex. 2. Investigate the cases /=i, and / = o. 



INDEX. 



PAGE 



Affini ty ."•;•" S3 ' 77 

Analytic expression for a Steinerian transformation 187 

expression for tangent and polar 118 

formulation of the problem of linkages for collineations 244 

representation of central projection 49 

Angle mirror 171 

Anharmonic ratio . . . . 5 

An optical problem. 167 

Apollonius 92 

Applications; of perspective 86 

in mechanics 217 

Asymptotes of conies 104 

Axes of conies 104 

Axis of perspective 30 

Bicircular cubics 190 

Bobillier 136 

Brianchon 133, 175 

Brianchon point 135 

Brianchon's theorem 134 

Cajori 179 

Casey 80 

Cayley 42 

Cayley's theorem 137 

Center; of a conic 103 

of perspective 30 

Central projection. 45, 86, 105, 146 

special cases of 51 

Chasles 59, 92, 222 

Circular cubics 204 

261 



262 INDEX. 

PAGE 

Circular points at infinity 20, 4 1 

Classification of cubics 195 

Clebsch 203 

Collineation .- 45, 59 

Complete quadrilateral, the 26 

Conies as intersections of right cones 149 

Conic sections 36, 92 

Conies in mechanical drawing and perspective 140 

Conjugate diameters. . 103 

Construction; of a parabola, etc 143 

of an equilateral hyperbola, etc 145 

of conies 146 

of foci independent of central projection 107 

of projective pencils and ranges 30 

Continuous groups of projective transformations 66 

Cremona 16, 166, 168 

Cubic; with oval and serpentine 195, 209 

the simple 195, 211 

with an isolated point 196, 211 

nodal 196, 212 

cuspidal 196, 213 

Cubics 189 

Curve of the second order 23 

Curves of the second order and class 24, 93 

Curves of the third order 189 

Czuber 207 

Desargues 1 1, 45, 1 74 

Diameters of conies 103 

Dilatation 60 

Directrix of a conic no 

Discriminant 99 

Discussion of collineation 63 

Distance circle 45 

Disteli 185, 203 

Doehlemann 146, 257 

Double points of the transformation 9 

Duality 68 

Elation 56, 79 

Ellipse 99, 104, 140 

Emch, Hermann 242, 253 



INDEX. 263 

PAGH 

Enriques iii 

Equation of a conic in line coordinates 122 

Equilateral hyperbola 145 

Euclid 5 

Exercises and problems 25, 34, 57, 70, 90 

Existence of ellipse, hyperbola, parabola, and their foci 104 

Feuerbach circle 179 

Fiedler iii, 25, 45, 137, 175 

Focal properties of conies 109 

Foci; of a conic 104 

construction of 107 

Funicular polygon 217, 219 

General collineation : 59 

General construction of projective pencils and ranges 30 

General reciprocal transformation 126 

Geometric determination and discussion of collineation 63 

Geometric quantities and their signs 1 

Geometry of stresses in a plane 223 

Gergonne 68 

Groups of transformation n, 66 

Harmonic ratio 12 

Hart 242 

Hilbert 136 

Homologous triangles • 80 

Hyperbola 99, 145 

Identity of curves of the second order and class, and conies 93 

Intersection of conic and straight line 161 

Invariant elements 68 

Inversor pantograph 248 

Involution n, 51 

Involution; of conjugate diameters 104 

of poles and polars 41 

of conjugate sections and stresses 224 

of stresses in nature 239 

of the pencil U+W 1 = o 172 

Involutoric pencils 18 

Involutoric transformations 12 



264 INDEX. 

PAGE 

Joachimsthal 26, 172 

Kempe 242 

Kempe's reversor 257 

Kirkmann's theorem 137 

Koenigs 242 

Koenigs' perspectivograph 258 

Kotter 179 

Lagrange 1 

Lame 22 

Lambert 45 

Laurent 2 

Leudesdorf 80 

Levy 229 

Lie 11, 57, 68 

Linear deformation 236 

Linear transformation 61, 244 

of a curve of the second order 94 

Maclaurin's theorem 35 

McCormack 1 

Mechanical drawing 140 

Menachmus 92 

Metric properties of the involution of stresses 229 

Minchin 222 

Moebius 3» 6 » 7, 53 

Monge 45> M9 

Newson 34, 68 

Newton 196 

Newton's theorem 35 

Oblique axial symmetry 56 

Optical problem 167 

Orthogonal axial symmetry 56 

Orthographic projection. 73 

Osculating circle of a conic 159 

Pantograph; Inversor, Sylvester's 248, 249 

Scheiner's 250 

Pappus 16, 1 ic 



INDEX. 265 



PAGE 



Parabola 09, 104, 143 

campaniformis cum ovali 195 

Pura 195 

puncta 196 

nodata 196 

cuspidata 196 

Parallel projection of conies 146 

Pascal 133 

Pascal line 135 

Peaucellier '. 242 

Peaucellier's inversor , 246 

Pencils and ranges of conies 172 

Pencils of rays , 15, 18 

Perspective 257 

Perspective; between any two given conies 151 

of a square 86 

of a circle 88 

Perspective pencils and ranges 29 

Perspectivograph, Koenigs 258 

Picard 71 

Plucker 68 

Poincare 5 

Polar and tangent 1 13 

Polar involution ; of the circle 38 

of conies 102 

Polar reciprocity 131 

Polars of a pencil, of conies. 172 

Polar systems - . 123 

Pole, polar of a circle 38 

Poncelet , 7, 36, 45, 68, 92, 149, *75 

Poncelet's principle of continuity 149 

Poncelet's problem 164 

Poudra 45 

Principle of duality 68 

Problem; in graphic statics 219 

in optics 167 

Problems of the second order 161 

Products of pencils and ranges of conies 180 

Products of projective pencils and ranges 22 

Projection; central. 45, 86 

orthographic 73 

Projective groups of transformation 11 



266 INDEX. 



FAGE 



Projective properties of the circle 35 

ranges and pencils 5, 15, 30 

theorems, statical proofs of 220 

transformations of the plane 59 

transformations of the points of a straight line 5 

Quadrilateral 26 

Quadruple, Steinerian 203 

Rabattement 77 

Realization of collineations by linkages 242 

Reciprocal polars 123 

Reciprocal transformation 123, 126 

Rectangular pair of an involution 19 

Reye 5, l6 , 9 2 

Reve's configuration 86 

Ritter, H , 257 

Ritter, W 228, 229 

Roberts 242 

Rotation 59 

Rotator 251 

Salmon 25, 135 

Scheiner's pantograph 250 

Self-polar triangle 42, 103 

Similitude 52 

Special cases of central projection 51 

Special constructions of conies by central projection and parallel projection 146 

Statical proofs ot some projective theorems 220 

Steiner 5, 22, 92, 185, 203 

Steinerian quadruple 203 

Steinerian transformation 185 

Steiner's theorem: 137 

Stress ellipse 229 

Stresses in a plane 223 

Sylvester 242 

Sylvester's pantograph. 249 

Symmetry 56 

Tangent and polar 118 

Tangents from a point to a conic 163 

Taylor 45 



INDEX. 267 



PAGE 



Theorems of Pascal and Brianchon 133 

Theory of conies 92 

Theory of reciprocal polars 123 

Traces; of a line • 75 

of a plane 76 

Transformation; hyperbolic, elliptic, parabolic 9, 13 

projective 5, 59 

linear 61, 244 

groups of 11, 66 

Translation 59 

Translators 252 

Various methods of generating a circular cubic 204 

Veronese hi 

Von Staudt iii, 5, 92 

Wiener , iii, 45 

Wurf 5 



<IJ\J- y 



